Tangents and Circle Problems

Competitive Mathematics · SSC CGL

3 Part30 worked examples135 practice questions

A tangent is a straight line that touches a circle at exactly one point; that point is the point of contact. A line that cuts the circle at two points is a secant. In the figure below a tangent rolls round the circle — the point of contact keeps changing, but the angle between the tangent and the radius to that point stays $9 0^{\circ}$ all the time.

◭ Diagram

Figure: the radius (blue) and the tangent (pink) turn together and the angle between them stays

9 0^{\circ}

Why is it $9 0^{\circ}$ ?

Take any point $Q$ on the tangent other than the point of contact $T$ . The line meets the circle only at $T$ , so $Q$ lies outside the circle, i.e. $OQ > OT$ (the radius). So $T$ is the point of the line nearest to the centre $O$ . The shortest distance from a point to a line is the perpendicular, hence $OT$ is perpendicular to the tangent.

What is in this chapter

Fact	Formula / rule	Where
Tangent ⟂ radius	$∠ OTP = 9 0^{\circ}$	Tangents from a point
Two tangents from a point are equal	$P A = PB = d^{2} - r^{2}$	Tangents from a point
Tangent and secant	$P T^{2} = P A \cdot PB$	Product rule
Alternate segment	tangent-chord angle = angle in the alternate segment	Alternate segment
Incircle	$r = \frac{Δ}{s}, x = s - a$	Circles inside shapes
Circumcircle and Euler	$O I^{2} = R (R - 2 r)$	Circles inside shapes

Letters in the table: $O$ = centre of the circle, $r$ = radius, $d$ = distance of the outside point $P$ from the centre, $T, A, B$ = points of contact or of cutting, $Δ$ = area of the triangle, $s$ = semi-perimeter, $a$ = the opposite side, $R$ = circumradius, $O I$ = distance between circumcentre and incentre.

Next chapterNext: common tangents of two circles, touching circles, the third circle, concentric circles and the angle-bisector questions are in the next chapter, Two Circles and Touching Circles.

Before you startYou should already know: Pythagoras, similar triangles (AA), the angle sum of a triangle, central angle = 2 × inscribed angle, the angle in a semicircle is

9 0^{\circ}

, opposite angles of a cyclic quadrilateral add to

18 0^{\circ}

(earlier chapters). They are not re-taught here — only used.

HabitFirst job in every question: draw the figure and join the point of contact to the centre. The moment you see a tangent a right-angled triangle appears, and half the questions in this chapter are just solving that right triangle.

Tangents from an outside point

Length · product · angle

Length of tangents from a point outside a circle, the chord of contact, the tangent-secant product and the alternate segment theorem.

Tangent from an outside point

Length · kite · angles

Two equal tangents from an outside point, $d^{2} - r^{2}$ and the angles of the kite.

The idea

From a point $P$ outside a circle exactly two tangents can be drawn: $P A$ and $PB$ ( $A, B$ the points of contact). Let $O$ be the centre, $d = OP$ and $r$ the radius. Since $O A ⊥ P A$ , triangle $O A P$ is right-angled at $A$ with hypotenuse $OP$ , so the length is $P A = d^{2} - r^{2}$ .

Why are the two tangents equal? In triangles $O A P$ and $OBP$ : $O A = OB$ (radii), $OP$ is common and each has a $9 0^{\circ}$ angle. Hypotenuse and one side are equal, so the triangles are congruent (RHS) and $P A = PB$ . The same fact makes $OP$ bisect both $∠ A PB$ and $∠ A OB$ . Quadrilateral $O A PB$ is a kite with two right angles.

◭ Diagram

Figure: the pink tangents are equal (

t

, green ticks), the blue radii make

9 0^{\circ}

A

and

B

, and the dashed green

OP

bisects both angles.

∑ Formula

t = P A = PB = d^{2} - r^{2}

Length of a tangent from an outside point

$t$: length of the tangent $P A = PB$
$d$: distance $OP$ of the outside point from the centre
$r$: radius of the circle

∑ Formula

∠ A PB + ∠ A OB = 18 0^{\circ}

What is left of the kite

O A PB

after its two right angles

$∠ A PB$: angle between the two tangents (at the outside point)
$∠ A OB$: angle the two points of contact make at the centre

If $∠ A PB = x$	then
$∠ A OB$	$18 0^{\circ} - x$
$∠ P A B = ∠ PB A$	$9 0^{\circ} - \frac{x}{2}$
$∠ O A B = ∠ OB A$	$\frac{x}{2}$
$∠ A PO$	$\frac{x}{2}$
Area of the kite	$r \cdot t$

Letters in the table: $x$ = the angle $∠ A PB$ between the tangents; $r$ = radius; $t$ = length of the tangent. Area: the kite is two right triangles $O A P$ , $OBP$ , each of area $\frac{1}{2} r t$ .

Spot itRecognise: “length of a tangent from an outside point” →

d^{2} - r^{2}

. “Shortest distance from the point to the circle” →

d - r

(and longest

d + r

). “Angle between the tangents given, angle at the centre asked” → subtract from

18 0^{\circ}

. Keep in mind

3 - 4 - 5

5 - 12 - 13

8 - 15 - 17

7 - 24 - 25

—

r

t

d

often sit in them.

Draw: centre $O$ , outside point $P$ , points of contact $A$ (and $B$ ). Join $O A$ and mark the right angle at $A$ .
From the two known quantities among $r$ , $t$ , $d$ find the third in the right triangle $O A P$ .
If angles are asked use the kite rules: $P A = PB$ gives an isosceles triangle, and $∠ P + ∠ O = 18 0^{\circ}$ .

◭ Diagram

Figure for the first example:

r = 9

d = 15

t = 12

(to scale).

Example 1 · Basic

A circle has radius 9 cm. A tangent is drawn from a point 15 cm from the centre. Find its length.

Solution, step by step

1The radius $O A$ to the point of contact is perpendicular to the tangent, so triangle $O A P$ is right-angled at $A$ . Its hypotenuse is the distance of the outside point, $OP = 15$ .
2By Pythagoras $P A^{2} = O P^{2} - O A^{2} = 225 - 81 = 144$ .
3So $P A = 144 = 12$ cm. (It is $9 - 12 - 15$ , three times $3 - 4 - 5$ .)

Answer 12 cm

The hypotenuse is always

d

— it is

d^{2} - r^{2}

, never the other way round.

◭ Diagram

Figure with the given measures.

Example 2 · Standard

From an outside point

P

tangents

P A

and

PB

are drawn,

O

is the centre and

∠ A PB = 6 4^{\circ}

. Find

∠ A OB

∠ P A B

and

∠ O A B

Solution, step by step

1In quadrilateral $O A PB$ the angles at $A$ and $B$ are $9 0^{\circ}$ , so $∠ A OB = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 4^{\circ} = 11 6^{\circ}$ .
2 $P A = PB$ , so in triangle $P A B$ we get $∠ P A B = ∠ PB A = \frac{18 0 ^{\circ} - 6 4 ^{\circ}}{2} = 5 8^{\circ}$ .
3 $O A ⊥ P A$ , so $∠ O A B = 9 0^{\circ} - ∠ P A B = 9 0^{\circ} - 5 8^{\circ} = 3 2^{\circ}$ . (Check: it equals $\frac{6 4 ^{\circ}}{2} = 3 2^{\circ}$ ✓)

Answer $11 6^{\circ}$ , $5 8^{\circ}$ , $3 2^{\circ}$

Direct rule: if

∠ P = x

then

∠ A OB = 18 0^{\circ} - x

∠ P A B = 9 0^{\circ} - \frac{x}{2}

∠ O A B = \frac{x}{2}

◭ Diagram

Figure with the given measures.

Example 3 · Standard

The tangent from an outside point

P

is 24 cm long and the nearest point of the circle is 18 cm from

P

. Find the radius.

Solution, step by step

1The point of the circle nearest to $P$ lies on the line to the centre, so $d = OP = r + 18$ .
2In the right triangle $t^{2} = d^{2} - r^{2}$ : $2 4^{2} = (r + 18)^{2} - r^{2}$ , i.e. $576 = 36 r + 324$ .
3 $r^{2}$ cancels, so $36 r = 252$ and $r = 7$ cm. Check: $d = 25$ , $625 - 49 = 24$ ✓.

Answer 7 cm

When you expand

(r + 18)^{2}

the

r^{2}

cancels, so the equation stays linear.

◭ Diagram

Figure for the last example:

∠ A PB = 4 0^{\circ}

Example 4 · Advanced

P A

and

PB

are tangents to a circle (

A, B

points of contact) and

∠ A PB = 4 0^{\circ}

. Points

C

and

D

are taken on the major and the minor arc

A B

. Find

∠ A CB

and

∠ A D B

Solution, step by step

1Kite rule: $∠ A OB = 18 0^{\circ} - 4 0^{\circ} = 14 0^{\circ}$ ( $O$ the centre).
2Any point $C$ on the major arc sees the chord $A B$ at an inscribed angle, half the central angle: $∠ A CB = \frac{14 0 ^{\circ}}{2} = 7 0^{\circ}$ .
3 $A CB D$ is a cyclic quadrilateral, so the opposite angles $C$ and $D$ add to $18 0^{\circ}$ : $∠ A D B = 18 0^{\circ} - 7 0^{\circ} = 11 0^{\circ}$ .

Answer $7 0^{\circ}$ and $11 0^{\circ}$

Direct formula: on the major arc

∠ A CB = 9 0^{\circ} - \frac{∠ P}{2}

; on the minor arc

9 0^{\circ} + \frac{∠ P}{2}

Practice — try yourself

BasicTry it yourself first

A tangent is drawn to a circle of radius 5 cm from a point 13 cm away from the centre. Find the length of the tangent (cm).

BasicTry it yourself first

A circular pond has radius 12 m. A person stands 20 m from its centre and walks straight along a tangent line to the point where it touches the bank. How many metres does the person walk?

BasicTry it yourself first

A circle has radius 12 cm and the tangent drawn from an outside point

P

is 16 cm long. Find the distance of

P

from the centre (cm).

StandardTry it yourself first

A tangent from a point 25 cm from the centre of a circle is 24 cm long. Find the diameter of the circle (cm).

BasicTry it yourself first

P A

and

PB

are tangents to a circle (

A, B

points of contact,

O

the centre) and

∠ A PB = 4 8^{\circ}

. Find

∠ A OB

StandardTry it yourself first

From an outside point

P

tangents

P A

PB

are drawn to a circle and

∠ A PB = 7 0^{\circ}

. Find

∠ P A B

StandardTry it yourself first

P A

PB

are tangents,

O

is the centre and

∠ O A B = 2 2^{\circ}

. Find

∠ A PB

StandardTry it yourself first

A circle has radius 15 cm and tangents

P A

PB

are drawn from a point

P

at 25 cm from the centre. Find the perimeter of quadrilateral

O A PB

(cm).

AdvancedTry it yourself first

The tangent

P A

is 40 cm long and

PO

cuts the circle at

C

with

PC = 16

cm (

O

the centre). Find the radius (cm).

AdvancedTry it yourself first

The tangents

P A

PB

have length

r 3

where

r

is the radius of the circle. Find

∠ A PB

Chord of contact

A B = \frac{2 r t}{d}

The chord joining the two points of contact: its length, its distance from the centre and the right triangles it makes.

The idea

The chord $A B$ joining the points of contact of the two tangents from $P$ is the chord of contact. The kite $O A PB$ is symmetric about its diagonal $OP$ , so $OP$ cuts $A B$ at right angles and in the middle. Call the meeting point $M$ .

Look at the right triangle $O A P$ (right angle at $A$ ): $A M$ is perpendicular to its hypotenuse $OP$ , i.e. the altitude on the hypotenuse. It has three well-known rules: altitude $= \frac{leg \times leg}{hypotenuse}$ , and each leg squared $=$ its piece $\times$ hypotenuse. All the formulas below come from these.

◭ Diagram

Figure: the saffron chord of contact

A B

meets the green line

OP

M

at a right angle and

A M = BM = a

∑ Formula

A B = \frac{2 r t}{d} OM = \frac{r ^{2}}{d} PM = \frac{t ^{2}}{d}

The chord of contact and the two pieces of

OP

$A B$: length of the chord of contact
$r$: radius
$t$: tangent length $d^{2} - r^{2}$
$d$: $OP$ , distance of the outside point from the centre
$OM, PM$: the two pieces of $OP$ made by the chord ( $OM + PM = d$ )

ReverseReverse question: if the chord

A B = 2 a

and

r

are given then

OM = r^{2} - a^{2}

, then

t = \frac{a r}{OM}

and

d = \frac{r ^{2}}{OM}

(from

△ OM A \sim △ O A P

). Check:

OM + PM = d

From the two known among $r$ , $d$ , $t$ find the third ( $t^{2} = d^{2} - r^{2}$ ).
Apply $A B = \frac{2 r t}{d}$ ; for the distance of the chord from the centre use $OM = \frac{r ^{2}}{d}$ .
If the chord is given take its half $a$ , find $OM = r^{2} - a^{2}$ , then $t$ and $d$ by similarity.

◭ Diagram

Figure with the given measures.

Example 5 · Basic

A circle has radius 9 cm and tangents are drawn from a point

P

15 cm from the centre. Find the length of the chord of contact and its distance from the centre.

Solution, step by step

1Tangent length $t = 1 5^{2} - 9^{2} = 144 = 12$ cm.
2The altitude of triangle $O A P$ on the hypotenuse is $A M = \frac{r \times t}{d} = \frac{9 \times 12}{15} = 7.2$ cm, so the chord $A B = 2 \times 7.2 = 14.4$ cm.
3Distance from the centre $OM = \frac{r ^{2}}{d} = \frac{81}{15} = 5.4$ cm. Check: $O M^{2} + A M^{2} = 29.16 + 51.84 = 81 = r^{2}$ ✓.

Answer $A B = 14.4$ cm, $OM = 5.4$ cm

You can always check:

O M^{2} + A M^{2} = r^{2}

◭ Diagram

Figure with the given measures.

Example 6 · Standard

A chord of length 24 cm is drawn in a circle of radius 20 cm. The tangents at its ends meet outside at

P

. Find the length of each tangent and

OP

Solution, step by step

1The tangents at the ends of the chord are the tangents from $P$ , so this is the chord of contact. Half of it is $a = 12$ and $OM = 2 0^{2} - 1 2^{2} = 256 = 16$ cm.
2From similar triangles $OM A$ and $O A P$ : $\frac{A P}{A M} = \frac{O A}{OM}$ , i.e. $t = \frac{12 \times 20}{16} = 15$ cm.
3From $O A^{2} = OM \cdot OP$ : $OP = \frac{400}{16} = 25$ cm. Check: $2 5^{2} - 2 0^{2} = 15$ ✓.

Answer Tangent 15 cm, $OP = 25$ cm

Here

OM + PM = 16 + 9 = 25

✓ (

PM = \frac{t ^{2}}{d} = 9

◭ Diagram

Figure with the given measures.

Example 7 · Advanced

From an outside point

P

tangents

P A = PB = 40

cm are drawn to a circle and the chord of contact

A B = 48

cm. Find the radius.

Solution, step by step

1From $A B = \frac{2 r t}{d}$ : $48 = \frac{2 r \times 40}{d}$ , i.e. $d = \frac{80 r}{48} = \frac{5 r}{3}$ .
2Also $d^{2} = r^{2} + t^{2}$ : $\frac{25 r ^{2}}{9} = r^{2} + 1600$ , so $\frac{16 r ^{2}}{9} = 1600$ and $r^{2} = 900$ .
3 $r = 30$ cm (then $d = 50$ ). Check: $\frac{2 \times 30 \times 40}{50} = 48$ ✓.

Answer 30 cm

Two equations, two unknowns (

r

and

d

): write

d

in terms of

r

from one and put it in the other.

Practice — try yourself

BasicTry it yourself first

A circle has radius 15 cm. Tangents

P A

PB

are drawn from a point

P

at 25 cm from the centre. Find the length of the chord of contact

A B

(cm).

BasicTry it yourself first

A circle has radius 24 cm. Tangents

P A

PB

are drawn from a point

P

at 25 cm from the centre. Find the length of the chord of contact

A B

(cm).

BasicTry it yourself first

Tangents are drawn to a circle of radius 12 cm (centre

O

) from an outside point

P

with

OP = 20

cm. Find the distance of the chord of contact from the centre (cm).

StandardTry it yourself first

A circle has radius 13 cm and the chord of contact of the two tangents from an outside point is 24 cm. Find the length of each tangent (cm).

StandardTry it yourself first

Tangents

P A = PB = 20

cm are drawn to a circle of radius 15 cm from a point

P

. Find the area of triangle

P A B

(sq cm).

StandardTry it yourself first

Tangents are drawn to a circle of radius 20 cm (centre

O

) from an outside point

P

with

OP = 25

cm. Find the distance of the chord of contact from the centre (cm).

AdvancedTry it yourself first

Tangents are drawn from a point

P

, 20 cm from the centre

O

, to a circle of radius 16 cm. Find the distance of the chord of contact from

P

(cm), i.e.

PM

StandardTry it yourself first

A circle has radius 21 cm. Tangents

P A

PB

are drawn from a point

P

at 35 cm from the centre. Find the length of the chord of contact

A B

(cm).

StandardTry it yourself first

A circle has radius 15 cm and the chord of contact of the two tangents from an outside point is 18 cm. Find the length of each tangent (cm).

AdvancedTry it yourself first

Tangents are drawn to a circle of radius 24 cm (centre

O

) from an outside point

P

with

OP = 30

cm. Find the distance of the chord of contact from the centre (cm).

The triangle cut off by a third tangent

Perimeter

= 2 t

A third tangent cutting two tangents: perimeter of the cut-off triangle, the angle at the centre and the shortest cut.

The idea

Draw the tangents $A B$ and $A C$ from $A$ ( $B, C$ points of contact). Now draw one more tangent at a point $P$ of the arc $BC$ , meeting $A B$ at $M$ and $A C$ at $N$ . $M$ is also an outside point, so its two tangents to the circle are equal: $MB = MP$ . Likewise $NC = NP$ .

The perimeter of triangle $A MN$ is $A M + MN + N A = A M + (MP + PN) + N A$ . Replace $MP$ by $MB$ and $PN$ by $NC$ : $= (A M + MB) + (NC + N A) = A B + A C$ . So the perimeter $= 2 \times A B$ wherever $P$ is!

◭ Diagram

Figure: pieces with the same green ticks are equal (

MB = MP

two ticks,

NC = NP

three ticks); the saffron triangle

A MN

has perimeter

A B + A C = 2 t

∑ Formula

A M + MN + N A = 2 t ∠ MON = 9 0^{\circ} - \frac{∠ A}{2}

Perimeter of the cut-off triangle and the angle at the centre

$t$: tangent length $A B = A C$ from the outside point $A$
$∠ A$: angle between the two tangents
$∠ MON$: angle made at the centre by $OM$ and $ON$

∑ Formula

M N_{m i n} = \frac{2 r ( d - r )}{t}

When the third tangent is drawn at the point where

O A

meets the circle (the shortest cut)

$r$: radius
$d$: $O A$
$t$: $d^{2} - r^{2}$

WhyWhy

9 0^{\circ} - \frac{A}{2}

: the circle touches the side

MN

and the extended sides

A M

A N

of triangle

A MN

, so

O

is its excentre, and at an excentre the opposite side subtends

9 0^{\circ} - \frac{A}{2}

. In the shortest cut

MN ⊥ O A

and

MP = PN

, so the similar triangles

A MP \sim A OB

help.

Spot the equal tangent pairs: two from $A$ , two from $M$ ( $MB, MP$ ), two from $N$ ( $NC, NP$ ).
Write the perimeter of the cut-off triangle $= A B + A C$ and subtract the known sides.
If an angle is asked use $9 0^{\circ} - \frac{A}{2}$ ; for the shortest cut use $MN = \frac{2 r ( d - r )}{t}$ .

◭ Diagram

Figure with the given measures.

Example 8 · Basic

The tangents from an outside point

A

are

A B = A C = 18

cm. A third tangent cuts

A B

A C

M

N

with

A M = 9

cm and

A N = 11

cm. Find

MN

Solution, step by step

1From $M$ two tangents to the circle: $MB$ and $MP$ — equal. From $N$ : $NC = NP$ . So $MN = MP + PN = MB + NC$ .
2 $MB = A B - A M = 18 - 9 = 9$ and $NC = A C - A N = 18 - 11 = 7$ .
3 $MN = 9 + 7 = 16$ cm. (Other way: perimeter $= 36$ , $36 - 9 - 11 = 16$ ✓)

Answer 16 cm

Both ways are the same fact — use whichever feels easier.

◭ Diagram

Figure with the given measures.

Example 9 · Standard

A B

A C

are tangents to a circle and

∠ B A C = 5 6^{\circ}

. A third tangent

MN

cuts off triangle

A MN

(

O

the centre). Find

∠ MON

Solution, step by step

1The circle touches $MN$ and the extended lines $A M$ , $A N$ , so $O$ is the excentre of triangle $A MN$ .
2 $OM$ and $ON$ bisect the exterior angles. In the triangle $∠ M + ∠ N = 18 0^{\circ} - 5 6^{\circ} = 12 4^{\circ}$ , the two exterior angles add to $36 0^{\circ} - 12 4^{\circ} = 23 6^{\circ}$ , and half of it is $11 8^{\circ}$ .
3In $△ MON$ : $∠ MON = 18 0^{\circ} - 11 8^{\circ} = 6 2^{\circ}$ . Check: $9 0^{\circ} - \frac{5 6 ^{\circ}}{2} = 6 2^{\circ}$ ✓.

Answer $6 2^{\circ}$

Instead of the long way remember:

∠ MON = 9 0^{\circ} - \frac{A}{2}

◭ Diagram

Figure for the third example: the proportions of

r = 6

d = 10

(

t = 8

);

MN ⊥ O A

Example 10 · Advanced

A circle has radius 6 cm and the outside point

A

is 10 cm from its centre. Tangents are drawn from

A

. The tangent at

P

, the point of the circle nearest to

A

, cuts them at

M

and

N

. Find

MN

and the perimeter of triangle

A MN

Solution, step by step

1 $t = 1 0^{2} - 6^{2} = 8$ cm; $A P = 10 - 6 = 4$ cm. The tangent at $P$ is perpendicular to $O A$ , so $∠ A PM = 9 0^{\circ}$ .
2Triangles $A MP$ and $A OB$ share angle $A$ and have a right angle each, so they are similar: $\frac{MP}{OB} = \frac{A P}{A B}$ , i.e. $MP = \frac{6 \times 4}{8} = 3$ cm.
3 $MN = 2 \times MP = 6$ cm ( $N$ is the same distance on the other side by symmetry). Perimeter $= 2 t = 16$ cm. Check: $A M = 4^{2} + 3^{2} = 5$ , $A N = 5$ , $5 + 6 + 5 = 16$ ✓.

Answer $MN = 6$ cm, perimeter 16 cm

Formula:

MN = \frac{2 r ( d - r )}{t} = \frac{2 \times 6 \times 4}{8} = 6

✓.

Practice — try yourself

BasicTry it yourself first

From an outside point

A

tangents

A B = A C = 20

cm are drawn to a circle. A third tangent cuts

A B

M

and

A C

N

, with

A M = 8

cm and

A N = 9

cm. Find

MN

(cm).

BasicTry it yourself first

From an outside point

A

tangents

A B = A C = 15

cm are drawn to a circle. A third tangent cuts

A B

M

and

A C

N

, with

A M = 6

cm and

A N = 7

cm. Find

MN

(cm).

BasicTry it yourself first

A B

A C

are tangents to a circle (centre

O

) and

∠ A = 6 0^{\circ}

. A third tangent

MN

cuts

A B

and

A C

M

N

(the circle lies outside triangle

A MN

). Find

∠ MON

StandardTry it yourself first

A B

A C

are tangents to a circle (centre

O

) and

∠ A = 7 4^{\circ}

. A third tangent

MN

cuts

A B

and

A C

M

N

(the circle lies outside triangle

A MN

). Find

∠ MON

StandardTry it yourself first

A circle has radius 6 cm and the outside point

A

is 10 cm from the centre. The tangent at

P

, the point of the circle nearest to

A

, cuts the two tangents from

A

M

and

N

. Find

MN

(cm).

AdvancedTry it yourself first

A circle has radius 8 cm and the outside point

A

is 17 cm from the centre. The tangent at

P

, the point of the circle nearest to

A

, cuts the two tangents from

A

M

and

N

. Find

MN

(cm).

StandardTry it yourself first

From an outside point

A

tangents

A B = A C = 30

cm are drawn to a circle. A third tangent cuts

A B

M

and

A C

N

, with

A M = 12

cm and

A N = 13

cm. Find

MN

(cm).

StandardTry it yourself first

A B

A C

are tangents to a circle (centre

O

) and

∠ A = 9 0^{\circ}

. A third tangent

MN

cuts

A B

and

A C

M

N

(the circle lies outside triangle

A MN

). Find

∠ MON

StandardTry it yourself first

A circle has radius 12 cm and the outside point

A

is 20 cm from the centre. The tangent at

P

, the point of the circle nearest to

A

, cuts the two tangents from

A

M

and

N

. Find

MN

(cm).

A special case: if $A B$ is a diameter with tangents at $A$ and $B$ , a tangent at a point $C$ of the arc cuts them at $D$ , $E$ with $D E = A D + BE$ and $r^{2} = A D \cdot BE$ (because $∠ D OE = 9 0^{\circ}$ ). The two questions below are on this.

StandardTry it yourself first

A B

is a diameter. A third tangent at a point of the arc cuts the tangents at

A

and

B

D

and

E

. If

A D = 3

cm and

BE = 12

cm, find the radius (cm).

AdvancedTry it yourself first

A B

is a diameter of a circle. Tangents are drawn at

A

and

B

and a third tangent at a point

C

of the arc cuts them at

D

and

E

. If

A D = 5

cm and

BE = 20

cm, find

D E

(cm).

Tangent-secant and chord products

P T^{2} = P A \cdot PB

The product rules for a tangent and a secant, two secants and chords meeting inside, and their link with the distance from the centre.

The idea

Draw a secant from an outside point $P$ cutting the circle at $A$ (near) and $B$ (far). Now turn the secant about $P$ : $A$ and $B$ come closer, and when the secant becomes a tangent they merge into one point $T$ . Then $P A = PB = PT$ and the product $P A \times PB$ becomes $PT \times PT = P T^{2}$ . The figure shows exactly this — and it tells us that $P A \cdot PB$ does not depend on the direction of the secant; it always equals $P T^{2}$ .

◭ Diagram

Figure: the green secant turns about

P

, the blue

A

and saffron

B

slide towards each other and finally meet at the pink point

T

Why? In triangles $PT A$ and $PBT$ angle $P$ is common and $∠ PT A = ∠ PBT$ (next section: the alternate segment theorem). Two equal angles make them similar: $\frac{PT}{PB} = \frac{P A}{PT}$ , i.e. $P T^{2} = P A \cdot PB$ . For two secants and for chords meeting inside, the same kind of similar triangles ( $P A D \sim PCB$ and $E A D \sim ECB$ ) give the same result.

Situation	Rule	How to remember
Tangent + secant (from outside)	$P T^{2} = P A \cdot PB$	tangent squared = near × far
Two secants (from outside)	$P A \cdot PB = PC \cdot P D$	near × far is equal for both
Two chords (meeting inside)	$E A \cdot EB = EC \cdot E D$	piece × piece is equal for both
Distance $d$ from the centre	$d^{2} - r^{2}$ / $r^{2} - d^{2}$	the product for an outside / inside point

Letters in the table: $P$ = the outside point; $PT$ = the tangent; $A, B$ and $C, D$ = cutting points of the secants (the first letter is the near one); $E$ = the inside point where chords meet; $r$ = radius, $d$ = distance of the point from the centre. Remember: always take the secant up to the far point ( $PB$ , not the chord $A B$ ).

◭ Diagram

Chords meeting inside: the product of the saffron pieces

a \cdot b

equals that of the pink pieces

c \cdot d

Spot itRecognise: “tangent and secant”, “two secants”, “chords meet inside”, or “a circle through

B, C

cutting

A B

A C

” → the product rule. The commonest question: three lengths given, the fourth wanted.

Pick the point ( $P$ or $E$ ) where the lines meet and decide whether it is outside or inside the circle.
For each line: from outside (near)×(far) (far = the whole secant), from inside (piece)×(piece). For a tangent use $PT \times PT$ .
Equate the products and solve for the unknown; then check whether the chord or the whole secant is asked.

◭ Diagram

Figure with the given measures.

Example 11 · Basic

From an outside point

P

a tangent

PT

and a secant

P A B

are drawn. If

P A = 4

cm and

A B = 5

cm, find

PT

Solution, step by step

1The whole secant runs from $P$ to the far point $B$ : $PB = P A + A B = 4 + 5 = 9$ cm.
2Tangent squared = near × far: $P T^{2} = P A \times PB = 4 \times 9 = 36$ .
3 $PT = 36 = 6$ cm.

Answer 6 cm

The commonest slip: taking

4 \times 5 = 20

. The product uses the whole

PB

◭ Diagram

Figure with the given measures.

Example 12 · Standard

Two chords

A B

and

C D

meet at an inside point

E

A E = 6

EB = 8

and

CE = 4

cm. Find

E D

and the whole chord

C D

Solution, step by step

1For chords meeting inside, the products of the pieces are equal: $A E \cdot EB = CE \cdot E D$ .
2 $6 \times 8 = 4 \times E D$ , i.e. $48 = 4 \times E D$ and $E D = 12$ cm.
3The whole chord $C D = CE + E D = 4 + 12 = 16$ cm. (Chord $A B = 14$ cm; no chord can exceed the diameter, so here the diameter is $\geq 16$ ✓.)

Answer $E D = 12$ cm, $C D = 16$ cm

Pieces are multiplied, not whole chords.

◭ Diagram

Figure with the given measures.

Example 13 · Advanced

A point

E

inside a circle of radius 13 cm is 5 cm from the centre. Find the shortest chord through

E

, and the other piece of a chord through

E

if one piece is 9 cm.

Solution, step by step

1For every chord through $E$ the product of its two pieces is the same: $r^{2} - d^{2} = 169 - 25 = 144$ .
2The shortest chord is perpendicular to $OE$ and is halved at $E$ , so each piece $x$ has $x \cdot x = 144$ , $x = 12$ and the chord $= 24$ cm.
3For the chord with one piece 9 the other piece is $\frac{144}{9} = 16$ cm (the whole chord is 25 cm, less than the diameter 26 ✓).

Answer Shortest chord 24 cm; other piece 16 cm

The longest chord through

E

is the diameter (pieces

13 + 5 = 18

and

13 - 5 = 8

, product

144

✓).

◭ Diagram

Figure with the given measures.

Example 14 · Advanced

Two circles intersect at

A

and

B

. The line

B A

extended meets the common tangent

PQ

M

. If

M A = 4

cm and

A B = 5

cm, find

PQ

Solution, step by step

1From $M$ the first circle has tangent $MP$ and secant $M A B$ , so $M P^{2} = M A \cdot MB$ . For the second circle $M Q^{2} = M A \cdot MB$ .
2So $MP = MQ$ : the line of the common chord bisects the common tangent.
3 $MB = 4 + 5 = 9$ , $M P^{2} = 4 \times 9 = 36$ , $MP = 6$ and $PQ = 2 \times 6 = 12$ cm.

Answer 12 cm

This trick hides in many questions: on seeing “common chord produced” write

MP = MQ

Practice — try yourself

BasicTry it yourself first

From a point

P

a tangent

PT

and a secant

P A B

are drawn to a circle (

A

is the nearer cutting point). If

P A = 9

cm and

A B = 7

cm, find

PT

(cm).

BasicTry it yourself first

The tangent from an outside point

P

PT = 12

cm. A secant through

P

cuts the circle at

A

(near) and

B

(far) with

P A = 8

cm. How long is the chord

A B

(cm)?

StandardTry it yourself first

Two secants are drawn from an outside point

P

: the first cuts the circle at

A

B

(

P A = 5

PB = 12

) and the second at

C

D

. If

PC = 6

, find the chord

C D

(cm).

StandardTry it yourself first

Chords

A B

and

C D

of a circle meet inside at

E

. If

A E = 9

EB = 8

and

CE = 12

(cm), find the length of chord

C D

(cm).

StandardTry it yourself first

A secant from a point

P

, 17 cm from the centre of a circle of radius 8 cm, cuts the circle at

A

(near) and

B

(far). If

P A = 9

cm, find

PB

(cm).

StandardTry it yourself first

A chord is drawn through a point

E

inside a circle of radius 10 cm, 6 cm from the centre, and

E

divides it into a piece of 4 cm and another piece. Find the other piece (cm).

StandardTry it yourself first

In triangle

A BC

A B = 10

cm and

A C = 12

cm. A circle through

B

and

C

cuts

A B

D

and

A C

E

. If

A D = 6

cm, find

A E

(cm).

AdvancedTry it yourself first

Two circles intersect at

A

and

B

. The line

B A

extended meets their common tangent

PQ

M

. If

M A = 9

cm and

A B = 16

cm, find the tangent segment

PQ

(cm).

BasicTry it yourself first

From a point

P

a tangent

PT

and a secant

P A B

are drawn to a circle (

A

is the nearer cutting point). If

P A = 2

cm and

A B = 16

cm, find

PT

(cm).

StandardTry it yourself first

The tangent from an outside point

P

PT = 20

cm. A secant through

P

cuts the circle at

A

(near) and

B

(far) with

P A = 10

cm. How long is the chord

A B

(cm)?

StandardTry it yourself first

Two secants are drawn from an outside point

P

: the first cuts the circle at

A

B

(

P A = 4

PB = 15

) and the second at

C

D

. If

PC = 5

, find the chord

C D

(cm).

BasicTry it yourself first

Chords

A B

and

C D

of a circle meet inside at

E

. If

A E = 4

EB = 9

and

CE = 6

(cm), find the length of chord

C D

(cm).

AdvancedTry it yourself first

A chord is drawn through a point

E

inside a circle of radius 26 cm, 10 cm from the centre, and

E

divides it into a piece of 18 cm and another piece. Find the other piece (cm).

AdvancedTry it yourself first

Two circles intersect at

A

and

B

. The line

B A

extended meets their common tangent

PQ

M

. If

M A = 1

cm and

A B = 8

cm, find the tangent segment

PQ

(cm).

Alternate segment theorem

Tangent-chord angle

=

angle in the alternate segment

The angle between a tangent and a chord equals the angle in the alternate segment — the key to angle questions.

The idea

Draw the tangent at $A$ and a chord $A B$ from $A$ . The chord splits the circle into two segments. The angle $θ$ between the tangent and the chord falls on one side; take any point $C$ in the opposite (alternate) segment, then $∠ A CB = θ$ . In the figure $C$ runs along the arc: the angle never changes.

◭ Diagram

Figure: the saffron angle

θ

between the pink tangent and the blue chord

A B

; the point

C

runs along the major arc and the angle

∠ A CB

made by the green lines is always

θ

Why is it true?

The radius $O A$ is perpendicular to the tangent, so in triangle $O A B$ we get $∠ O A B = 9 0^{\circ} - θ$ . $O A = OB$ makes $∠ OB A$ the same, so at the centre $∠ A OB = 18 0^{\circ} - 2 (9 0^{\circ} - θ) = 2 θ$ . An inscribed angle is half the central angle, so any point on the major arc gives $∠ A CB = θ$ . (A point on the minor arc gives $18 0^{\circ} - θ$ — the cyclic quadrilateral.)

◭ Diagram

Static figure: tangent-chord angle

θ

, the angle

θ

in the alternate segment and

2 θ

at the centre.

∑ Formula

∠ (tangent, A B) = ∠ A CB = \frac{1}{2} ∠ A OB

The three forms of the theorem

$∠ (tangent, A B)$: angle between the tangent at $A$ and the chord $A B$ (on the side of arc $A B$ )
$∠ A CB$: angle made at a point $C$ of the opposite segment
$∠ A OB$: central angle of the chord $A B$

Spot itRecognise: a tangent is in the figure and a chord leaves the point of contact, an angle is given → write the same angle at a point on the other side of the chord at once. If the tangent is parallel to a chord, or tangents at both ends of a chord appear, isosceles triangles come (

P A = PB

∠ P A B = ∠ A CB

Identify the point of contact $A$ and the chord $A B$ leaving it.
Write the tangent-chord angle again as $∠ A CB$ at a point $C$ of the arc on the other side (or $2 \times$ at the centre).
Now find the rest with the triangle sum $18 0^{\circ}$ , a straight line $18 0^{\circ}$ or the cyclic quadrilateral.

◭ Diagram

Figure with the given measures.

Example 15 · Basic

A T

is the tangent at

A

A B

a chord and

∠ T A B = 5 2^{\circ}

. Find

∠ A CB

in the alternate segment and

∠ A OB

at the centre.

Solution, step by step

1Alternate segment theorem: the angle in the alternate segment equals the tangent-chord angle, so $∠ A CB = 5 2^{\circ}$ .
2The angle at the centre is twice the inscribed angle: $∠ A OB = 2 \times 5 2^{\circ} = 10 4^{\circ}$ .

Answer $5 2^{\circ}$ and $10 4^{\circ}$

Check: from the radius

∠ O A B = 9 0^{\circ} - 5 2^{\circ} = 3 8^{\circ}

, so

38 + 38 + 104 = 180

✓.

◭ Diagram

Figure with the given measures.

Example 16 · Standard

Triangle

A BC

is inscribed in a circle. The tangent

S A T

A

makes

∠ S A B = 3 5^{\circ}

and

∠ T A C = 4 5^{\circ}

. Find the three angles of the triangle.

Solution, step by step

1Between the chord $A B$ and the tangent is $3 5^{\circ}$ ; the angle in the alternate segment is $∠ A CB = 3 5^{\circ}$ .
2Similarly between the chord $A C$ and the tangent is $4 5^{\circ}$ , so $∠ A BC = 4 5^{\circ}$ .
3 $∠ B A C = 18 0^{\circ} - 3 5^{\circ} - 4 5^{\circ} = 10 0^{\circ}$ . Check: on the straight line at $A$ : $35 + 100 + 45 = 180$ ✓.

Answer $∠ A = 10 0^{\circ}$ , $∠ B = 4 5^{\circ}$ , $∠ C = 3 5^{\circ}$

The two angles at the tangent are angles at the other vertices:

∠ S A B \to C

∠ T A C \to B

(the one opposite the chord).

◭ Diagram

Figure with the given measures.

Example 17 · Standard

The tangents at

A

and

B

of a circle meet at

P

and

∠ A PB = 7 0^{\circ}

. Find

∠ A CB

for a point

C

on the major arc and

∠ A D B

for a point

D

on the minor arc.

Solution, step by step

1 $P A = PB$ , so in triangle $P A B$ : $∠ P A B = \frac{18 0 ^{\circ} - 7 0 ^{\circ}}{2} = 5 5^{\circ}$ .
2 $∠ P A B$ is a tangent-chord angle, so in the alternate segment (major arc) $∠ A CB = 5 5^{\circ}$ .
3On the minor arc $A CB D$ is a cyclic quadrilateral: $∠ A D B = 18 0^{\circ} - 5 5^{\circ} = 12 5^{\circ}$ .

Answer $5 5^{\circ}$ and $12 5^{\circ}$

The answer from the earlier method:

9 0^{\circ} - \frac{7 0 ^{\circ}}{2} = 5 5^{\circ}

✓ — both routes agree.

◭ Diagram

Figure with the given measures.

Example 18 · Advanced

A, B, C

are points on a circle. The tangent at

A

meets the chord

BC

produced at

T

(

C

lies between

B

and

T

). If

∠ A TC = 3 0^{\circ}

and

∠ C A T = 5 0^{\circ}

, find

∠ A CB

and

∠ BOC

Solution, step by step

1The angle between the chord $A C$ and the tangent is $∠ C A T = 5 0^{\circ}$ , so by the alternate segment $∠ A BC = 5 0^{\circ}$ .
2In triangle $A BT$ : $∠ B A T = 18 0^{\circ} - 5 0^{\circ} - 3 0^{\circ} = 10 0^{\circ}$ , and removing $∠ C A T = 5 0^{\circ}$ leaves $∠ B A C = 5 0^{\circ}$ .
3In triangle $A BC$ : $∠ A CB = 18 0^{\circ} - 5 0^{\circ} - 5 0^{\circ} = 8 0^{\circ}$ , and the central angle $∠ BOC = 2 \times ∠ B A C = 10 0^{\circ}$ .

Answer $∠ A CB = 8 0^{\circ}$ , $∠ BOC = 10 0^{\circ}$

Remember the order: alternate segment → triangle sum → central angle.

Practice — try yourself

BasicTry it yourself first

A T

is the tangent at

A

and

A B

a chord.

∠ T A B = 3 8^{\circ}

and

C

is a point on the arc of the alternate segment. Find

∠ A CB

BasicTry it yourself first

PQ

is the tangent at

A

to a circle,

A B

is a chord and

∠ P A B = 4 1^{\circ}

. Find

∠ A OB

at the centre

O

BasicTry it yourself first

An arc

A B

of a circle measures

13 0^{\circ}

. What is the angle between the tangent at

A

and the chord

A B

, on the side of the arc

A B

StandardTry it yourself first

Triangle

A BC

is inscribed in a circle. The tangent

S A T

A

makes

∠ S A B = 2 8^{\circ}

and

∠ T A C = 6 4^{\circ}

. Find

∠ B A C

StandardTry it yourself first

The tangent at

A

to the circumcircle of triangle

A BC

is parallel to

BC

and

∠ B A C = 4 0^{\circ}

. Find

∠ A BC

AdvancedTry it yourself first

A, B, C

are points on a circle. The tangent at

A

meets the chord

BC

produced at

T

(

C

lies between

B

and

T

). If

∠ A TC = 3 4^{\circ}

and

∠ C A T = 4 6^{\circ}

, find the central angle

∠ BOC

◭ Diagram

StandardTry it yourself first

Triangle

A BC

is inscribed in a circle. The tangent

S A T

A

makes

∠ S A B = 5 2^{\circ}

and

∠ T A C = 7 1^{\circ}

. Find

∠ B A C

BasicTry it yourself first

A T

is the tangent at

A

and

A B

a chord.

∠ T A B = 6 4^{\circ}

and

C

is a point on the arc of the alternate segment. Find

∠ A CB

StandardTry it yourself first

PQ

is the tangent at

A

to a circle,

A B

is a chord and

∠ P A B = 3 3^{\circ}

. Find

∠ A OB

at the centre

O

StandardTry it yourself first

An arc

A B

of a circle measures

9 8^{\circ}

. What is the angle between the tangent at

A

and the chord

A B

, on the side of the arc

A B

StandardTry it yourself first

The tangent at

A

to the circumcircle of triangle

A BC

is parallel to

BC

and

∠ B A C = 6 4^{\circ}

. Find

∠ A BC

◭ Diagram

AdvancedTry it yourself first

A, B, C

are points on a circle. The tangent at

A

meets the chord

BC

produced at

T

(

C

lies between

B

and

T

). If

∠ A TC = 2 4^{\circ}

and

∠ C A T = 3 8^{\circ}

, find the central angle

∠ BOC

Circles inside shapes

Incircle · circumcircle · sector

Circles inside quadrilaterals, triangles and sectors: inradius, circumradius, excircle and Euler's relation.

Tangential quadrilateral

A B + C D = BC + D A

A quadrilateral whose four sides all touch one circle: the opposite-sides rule and the trapezium cases.

The idea

All four sides of quadrilateral $A BC D$ touch a circle (the circle lies inside). From each vertex two tangents can be drawn to the circle and they are equal. Call these lengths $p, q, r, s$ for the vertices $A, B, C, D$ . Then the sides are $A B = p + q$ , $BC = q + r$ , $C D = r + s$ , $D A = s + p$ .

◭ Diagram

Figure: two pieces of the same colour are the equal tangents from one vertex. Adding opposite sides, each letter appears once.

The sums of opposite sides: $A B + C D = (p + q) + (r + s)$ and $BC + D A = (q + r) + (s + p)$ — both equal $p + q + r + s$ . So $A B + C D = BC + D A$ . The semi-perimeter is $A B + C D$ , and joining the centre to the four vertices gives four triangles of height $r$ , so area $= r \times$ semi-perimeter.

∑ Formula

A B + C D = BC + D A Area = r \cdot s

The two rules of a tangential quadrilateral

$A B, BC, C D, D A$: the four sides
$r$: radius of the inside circle
$s$: semi-perimeter $= \frac{A B + BC + C D + D A}{2}$

Shape	When a circle fits	Useful fact
Rhombus / square	always	$r = \frac{d _{1} d _{2}}{4 side}$
Rectangle	only a square	$r = \frac{a}{2}$
Parallelogram	only a rhombus	—
Isosceles trapezium	$leg = \frac{a + b}{2}$	$h = ab, r = \frac{ab}{2}$
Right trapezium	$A B + C D = a + b$	$r = \frac{ab}{a + b}$

Letters in the table: $a, b$ = the two parallel sides of the trapezium; $h$ = height ( $= 2 r$ ); $d_{1}, d_{2}$ = diagonals of the rhombus; $a$ (in the rectangle row) = side of the square.

Spot itRecognise: “a circle touches all four sides” or “the quadrilateral circumscribes a circle” → start with

A B + C D = BC + D A

. With a trapezium the slant sides add up to the parallel sides; the height then comes from Pythagoras.

Notice that the circle touches all four sides and write $A B + C D = BC + D A$ .
Find the unknown side; for a trapezium the slant side is $\frac{a + b}{2}$ (isosceles) or use Pythagoras for the height.
For the radius or the area use $area = r \times s$ or $h = 2 r$ .

◭ Diagram

Figure with the given measures.

Example 19 · Basic

Quadrilateral

A BC D

circumscribes a circle.

A B = 14

BC = 9

C D = 11

cm. Find

A D

and the perimeter.

Solution, step by step

1The circle touches all four sides, so $A B + C D = BC + A D$ : $14 + 11 = 9 + A D$ .
2 $A D = 25 - 9 = 16$ cm.
3Perimeter $= 14 + 9 + 11 + 16 = 50$ cm, which is $2 \times 25$ (both pairs add to the same ✓).

Answer $A D = 16$ cm, perimeter 50 cm

Perimeter

= 2 \times (A B + C D)

— even without finding the side.

◭ Diagram

Figure with the given measures.

Example 20 · Standard

An isosceles trapezium has parallel sides 16 cm and 4 cm and a circle inside touches all four sides. Find the slant side, the height, the radius and the area.

Solution, step by step

1The slant sides add up to the parallel sides $= 16 + 4 = 20$ , and they are equal, so each is $10$ cm.
2Drop a perpendicular: it cuts $\frac{16 - 4}{2} = 6$ cm from the base, so the height is $h = 1 0^{2} - 6^{2} = 8$ cm (formula $ab = 64 = 8$ ✓).
3The circle touches both parallel sides, so $h = 2 r$ and $r = 4$ cm. Area $= \frac{1}{2} (16 + 4) \times 8 = 80$ sq cm. Check: $r \times s = 4 \times 20 = 80$ ✓.

Answer Slant side 10, height 8, $r = 4$ cm, area 80 sq cm

Check the area two ways —

\frac{1}{2} (a + b) h

and

r \times s

— they must agree.

◭ Diagram

Figure with the given measures.

Example 21 · Advanced

In right trapezium

A BC D

A B ⊥ A D

A B ⊥ BC

A D = 10

cm,

BC = 15

cm and a circle touches all four sides. Find

C D

A B

and the radius.

Solution, step by step

1Let $A B = h$ and $C D = c$ . Tangential quadrilateral: $h + c = 10 + 15 = 25$ , i.e. $c = 25 - h$ .
2Drop a perpendicular from $D$ to $BC$ : a right triangle with sides $h$ and $15 - 10 = 5$ , so $c^{2} = h^{2} + 25$ . Then $(25 - h)^{2} = h^{2} + 25$ , i.e. $625 - 50 h = 25$ and $h = 12$ .
3 $A B = 12$ cm, $C D = 25 - 12 = 13$ cm (5-12-13 ✓) and $r = \frac{12}{2} = 6$ cm. Formula check: $\frac{10 \times 15}{25} = 6$ ✓.

Answer $C D = 13$ cm, $A B = 12$ cm, $r = 6$ cm

h^{2}

cancels on both sides — the equation stays linear.

Practice — try yourself

BasicTry it yourself first

A circle inside quadrilateral

A BC D

touches all four sides. If

A B = 9

BC = 7

and

C D = 12

cm, find

A D

(cm).

BasicTry it yourself first

A circle inside quadrilateral

A BC D

touches all four sides. If

A B = 21

BC = 15

and

C D = 18

cm, find

A D

(cm).

StandardTry it yourself first

A quadrilateral is drawn round a circle (touching all four sides) with opposite sides

A B = 17

cm and

C D = 13

cm, and the radius is 6 cm. Find the area of the quadrilateral (sq cm).

StandardTry it yourself first

A trapezium has parallel sides 18 cm and 8 cm, its non-parallel sides are equal and a circle inside touches all four sides. Find the area of the trapezium (sq cm).

StandardTry it yourself first

In right trapezium

A BC D

A B ⊥ A D

A B ⊥ BC

A D = 6

cm and

BC = 12

cm. A circle inside touches all four sides. Find the radius (cm).

AdvancedTry it yourself first

In right trapezium

A BC D

A B ⊥ A D

A B ⊥ BC

A D = 12

cm and

BC = 36

cm. A circle inside touches all four sides. Find the radius (cm).

AdvancedTry it yourself first

The diagonals of a rhombus are 16 cm and 12 cm. Find the radius of the circle inside it that touches all four sides (cm).

◭ Diagram

BasicTry it yourself first

A circle inside quadrilateral

A BC D

touches all four sides. If

A B = 12

BC = 8

and

C D = 10

cm, find

A D

(cm).

StandardTry it yourself first

A trapezium has parallel sides 27 cm and 3 cm, its non-parallel sides are equal and a circle inside touches all four sides. Find the area of the trapezium (sq cm).

StandardTry it yourself first

In right trapezium

A BC D

A B ⊥ A D

A B ⊥ BC

A D = 8

cm and

BC = 24

cm. A circle inside touches all four sides. Find the radius (cm).

AdvancedTry it yourself first

A quadrilateral is drawn round a circle (touching all four sides) with opposite sides

A B = 29

cm and

C D = 21

cm, and the radius is 5 cm. Find the area of the quadrilateral (sq cm).

Incircle of a triangle

r = \frac{Δ}{s}

x = s - a

The inradius, the tangent lengths from the vertices ( $s - a$ ) and the right-triangle shortcut.

The idea

The incircle of a triangle touches all three sides; its centre $I$ (the incentre) is where the angle bisectors meet. In the figure the circle starts big and shrinks until it just touches all three sides.

◭ Diagram

Figure: the circle shrinks until it touches the sides. The two tangents from

A

are

x

(blue), from

B

are

y

(green), from

C

are

z

(saffron);

r

is the radius to a point of contact.

The two tangents from each vertex are equal, so the sides are $a = y + z$ , $b = z + x$ , $c = x + y$ . Add them: $a + b + c = 2 (x + y + z)$ , i.e. $x + y + z = s$ (the semi-perimeter). Hence $x = s - a$ , $y = s - b$ , $z = s - c$ — the tangent from $A$ is $s$ minus the side $a$ opposite to it.

Area: joining $I$ to the three vertices gives three triangles of height $r$ and bases $a, b, c$ : $Δ = \frac{1}{2} r a + \frac{1}{2} r b + \frac{1}{2} rc = r \cdot s$ . Hence $r = \frac{Δ}{s}$ .

∑ Formula

r = \frac{Δ}{s} x = s - a Δ = s (s - a) (s - b) (s - c)

Inradius, tangent lengths and Heron's formula

$r$: radius of the incircle
$Δ$: area of the triangle
$s$: semi-perimeter $\frac{a + b + c}{2}$
$x$: length from vertex $A$ to the points of contact (opposite $BC = a$ )
$a, b, c$: the sides $BC$ , $C A$ , $A B$

◭ Diagram

Right triangle: the green square (side

r

) sits at the right angle, so the hypotenuse

h = (p - r) + (q - r)

, i.e.

r = \frac{p + q - h}{2}

∑ Formula

r = \frac{p + q - h}{2} (right triangle) r = \frac{a 3}{6} (equilateral)

Two quick shortcuts

$p, q$: the two legs of the right triangle
$h$: hypotenuse
$a$: side of the equilateral triangle

Spot itRecognise: “incircle”, “largest circle inside”, “circle touching the sides” →

r = Δ/ s

. Side

=

sum of two tangent pieces (

a = y + z

) — when the pieces cut by the points of contact are given. At the incentre

∠ B I C = 9 0^{\circ} + \frac{A}{2}

Find $s$ , then the area (Heron, or $\frac{1}{2} pq$ for a right triangle).
$r = \frac{Δ}{s}$ . For the tangent pieces use $x = s - a$ , $y = s - b$ , $z = s - c$ .
If the pieces are given build the sides by adding: $a = y + z$ etc.

◭ Diagram

Figure with the given measures.

Example 22 · Basic

The sides of a triangle are 17, 25 and 28 cm. Find the inradius and the tangent length from each vertex.

Solution, step by step

1Semi-perimeter $s = \frac{17 + 25 + 28}{2} = 35$ . Heron: $Δ = 35 \times 18 \times 10 \times 7 = 44100 = 210$ sq cm.
2 $r = \frac{Δ}{s} = \frac{210}{35} = 6$ cm.
3Tangent pieces: from the vertex opposite 17 it is $35 - 17 = 18$ , opposite 25 it is $35 - 25 = 10$ , opposite 28 it is $35 - 28 = 7$ . Check: $10 + 7 = 17$ , $18 + 7 = 25$ , $18 + 10 = 28$ ✓.

Answer $r = 6$ cm; pieces 18, 10, 7 cm

The tangent piece comes from the opposite side: the vertex opposite the longest side gets the shortest piece.

◭ Diagram

Figure with the given measures.

Example 23 · Standard

A right triangle has sides 15, 20 and 25 cm. Find the inradius in two ways and the tangent length from the right-angle vertex.

Solution, step by step

1Shortcut: $r = \frac{15 + 20 - 25}{2} = 5$ cm.
2Long way: $Δ = \frac{1}{2} \times 15 \times 20 = 150$ , $s = 30$ , $r = \frac{150}{30} = 5$ cm ✓.
3The tangent from the right-angle vertex is $s - 25 = 5$ cm — equal to $r$ , because a square forms there (centre, two points of contact and the vertex).

Answer $r = 5$ cm; tangent 5 cm

When two routes agree the answer is safe.

◭ Diagram

Figure with the given measures.

Example 24 · Advanced

The incircle of triangle

A BC

touches

BC

D

with

B D = 6

cm,

D C = 8

cm and the inradius is 4 cm. Find

A B

and

A C

Solution, step by step

1Let the tangent from $A$ be $x$ . Then $A B = x + 6$ , $A C = x + 8$ , $BC = 14$ and $s = x + 14$ .
2Area two ways: $Δ = r \cdot s = 4 (x + 14)$ and by Heron $Δ = s \cdot x \cdot 6 \cdot 8 = (x + 14) \cdot 48 x$ (since $s - a = x$ , $s - b = 8$ , $s - c = 6$ ).
3Equate the squares: $16 (x + 14)^{2} = 48 x (x + 14)$ ; dividing by $(x + 14)$ : $16 (x + 14) = 48 x$ , $224 = 32 x$ and $x = 7$ .
4 $A B = 13$ cm, $A C = 15$ cm (and $BC = 14$ ). Check: $Δ = 84$ , $s = 21$ , $r = 4$ ✓.

Answer $A B = 13$ cm, $A C = 15$ cm

Equating

Δ = rs

with Heron is the standard way to find the unknown

x

Practice — try yourself

BasicTry it yourself first

The sides of a triangle are 13, 14 and 15 cm. Find the radius of its incircle (cm).

BasicTry it yourself first

The sides forming the right angle of a right triangle are 9 cm and 12 cm. Find the inradius (cm).

BasicTry it yourself first

The incircle of triangle

A BC

touches

BC

C A

A B

D

E

F

. If

A F = 3

B D = 5

and

CE = 6

cm, find the perimeter (cm).

StandardTry it yourself first

A triangular garden has sides 20 m, 13 m and 11 m. Find the radius (m) of the largest round fountain that fits in it, touching all three sides.

StandardTry it yourself first

In triangle

A BC

∠ A = 6 4^{\circ}

and

I

is its incentre. Find

∠ B I C

StandardTry it yourself first

The sides forming the right angle of a right triangle are 20 cm and 21 cm. Find the inradius (cm).

AdvancedTry it yourself first

Find the radius (cm) of the incircle of an equilateral triangle of side 12 cm.

StandardTry it yourself first

The sides of a triangle are 9, 10 and 17 cm. Find the radius of its incircle (cm).

◭ Diagram

StandardTry it yourself first

The incircle of triangle

A BC

touches

BC

C A

A B

D

E

F

. If

A F = 6

B D = 7

and

CE = 8

cm, find the perimeter (cm).

StandardTry it yourself first

The sides forming the right angle of a right triangle are 8 cm and 15 cm. Find the inradius (cm).

BasicTry it yourself first

In triangle

A BC

∠ A = 8 0^{\circ}

and

I

is its incentre. Find

∠ B I C

Circumcircle, excircle and Euler's relation

R = \frac{ab c}{4Δ}

O I^{2} = R (R - 2 r)

Circumradius, exradius, Euler's formula for the distance between circumcentre and incentre, and the relations among them.

The idea

The circle through all three vertices is the circumcircle (radius $R$ , centre $O$ ); the inside circle touching all three sides is the incircle (radius $r$ , centre $I$ ). The outside circle touching one side and the extended lines of the other two is an excircle (radius $r_{a}$ , $r_{b}$ , $r_{c}$ ).

◭ Diagram

Figure: circumcircle (outer,

R

) and incircle (inner,

r

); the saffron line is the distance

d = O I

between the centres. Always

d^{2} = R (R - 2 r)

∑ Formula

R = \frac{ab c}{4Δ} r_{a} = \frac{Δ}{s - a} O I^{2} = R (R - 2 r)

Circumradius, exradius and Euler's formula

$R$: radius of the circumcircle
$a, b, c$: the sides of the triangle
$Δ$: area
$s$: semi-perimeter
$r_{a}$: radius of the excircle touching side $a$
$O I$: distance between circumcentre and incentre

◭ Diagram

Figure: the violet excircle outside

BC

grows until it touches

BC

and the extended lines

A B

A C

; the tangent from

A

to it has length

s

(the semi-perimeter).

Triangle	$R$	$r$	$O I$
Right (hypotenuse $h$ , legs $p, q$ )	$\frac{h}{2}$	$\frac{p + q - h}{2}$	$R (R - 2 r)$
Equilateral (side $a$ )	$\frac{a}{3}$	$\frac{a}{2 3}$	$0$
Any triangle	$\frac{ab c}{4Δ}$	$\frac{Δ}{s}$	$R^{2} - 2 R r$

Letters in the table: $h$ = hypotenuse, $p, q$ = the legs, $a$ = side of the equilateral triangle. More relations: $R \geq 2 r$ (equality only for the equilateral triangle); $r_{a} + r_{b} + r_{c} = 4 R + r$ ; $\frac{1}{r _{a}} + \frac{1}{r _{b}} + \frac{1}{r _{c}} = \frac{1}{r}$ ; $r_{a} r_{b} r_{c} r = Δ^{2}$ .

RememberWhy

R \geq 2 r

O I^{2} = R (R - 2 r)

cannot be negative, so

R - 2 r \geq 0

. If a question gives

R < 2 r

no such triangle exists. Recognise: “distance between circumcentre and incentre” → Euler; “outside circle touching one side” →

\frac{Δ}{s - a}

From the sides find $s$ and $Δ$ (Heron).
Take the radius you need: $R = \frac{ab c}{4Δ}$ , $r = \frac{Δ}{s}$ or $r_{a} = \frac{Δ}{s - a}$ .
For the distance $O I$ find both $R$ and $r$ , then $O I^{2} = R (R - 2 r)$ .

Example 25 · Basic

The sides of a triangle are 13, 14, 15 cm. Find

R

r

and

O I

Solution, step by step

1 $s = 21$ , $Δ = 21 \times 8 \times 7 \times 6 = 84$ sq cm.
2 $r = \frac{84}{21} = 4$ cm and $R = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{2730}{336} = \frac{65}{8}$ cm $= 8.125$ .
3Euler: $O I^{2} = R (R - 2 r) = \frac{65}{8} \times (\frac{65}{8} - 8) = \frac{65}{8} \times \frac{1}{8} = \frac{65}{64}$ , so $O I = \frac{65}{8} \approx 1.01$ cm.

Answer $R = \frac{65}{8}$ , $r = 4$ , $O I = \frac{65}{8}$

O I

is tiny — in an ordinary triangle the two centres lie close together.

Example 26 · Standard

For the same triangle (13, 14, 15) find the radii of the three excircles and check

r_{a} + r_{b} + r_{c} = 4 R + r

. (

a = 14

b = 15

c = 13

)

Solution, step by step

1 $s = 21$ , $Δ = 84$ . $r_{a} = \frac{Δ}{s - a} = \frac{84}{21 - 14} = 12$ .
2 $r_{b} = \frac{84}{21 - 15} = 14$ and $r_{c} = \frac{84}{21 - 13} = 10.5$ cm.
3Check: $12 + 14 + 10.5 = 36.5$ and $4 R + r = 4 \times 8.125 + 4 = 36.5$ ✓.

Answer 12, 14 and 10.5 cm

The excircle on the shortest side is the biggest because

s - a

is the largest there; the formula divides by

s - a

◭ Diagram

Figure with the given measures.

Example 27 · Advanced

A triangle has circumradius 25 cm and the distance from circumcentre to incentre is 15 cm. Find the inradius and say why the triangle is not equilateral.

Solution, step by step

1Euler: $O I^{2} = R^{2} - 2 R r$ , so $225 = 625 - 50 r$ .
2 $50 r = 400$ and $r = 8$ cm.
3In an equilateral triangle $O I = 0$ and $R = 2 r$ ; here $O I = 15 \neq = 0$ and $2 r = 16 < 25 = R$ , so the triangle is not equilateral (but $R \geq 2 r$ ✓).

Answer $r = 8$ cm

Final check always:

R \geq 2 r

must hold.

Practice — try yourself

BasicTry it yourself first

An isosceles triangle has equal sides 10 cm and base 12 cm. Find the radius of its circumcircle (cm).

StandardTry it yourself first

An isosceles triangle has equal sides 15 cm and base 24 cm. Find the radius of its circumcircle (cm).

StandardTry it yourself first

The sides of a triangle are 15, 15, 24 cm. Find the distance

O I

(cm) between its circumcentre

O

and incentre

I

StandardTry it yourself first

The sides of a triangle are 10, 10, 12 cm. Find the distance

O I

(cm) between its circumcentre

O

and incentre

I

StandardTry it yourself first

In triangle

A BC

BC = 13

C A = 14

A B = 15

cm. Find the radius (cm) of the excircle opposite

A

(touching side

BC

AdvancedTry it yourself first

In triangle

A BC

BC = 5

C A = 12

A B = 13

cm. Find the radius (cm) of the excircle opposite

B

(touching side

C A

AdvancedTry it yourself first

A triangle has circumradius 12.5 cm and the distance between circumcentre and incentre is 7.5 cm. Find the inradius (cm).

BasicTry it yourself first

The legs of a right triangle are 20 cm and 21 cm. Find the radius of its circumcircle (cm).

StandardTry it yourself first

In triangle

A BC

BC = 13

C A = 14

A B = 15

cm. Find the radius (cm) of the excircle opposite

B

(touching side

C A

StandardTry it yourself first

A triangle has circumradius 10 cm and the distance between circumcentre and incentre is 8 cm. Find the inradius (cm).

AdvancedTry it yourself first

An isosceles triangle has equal sides 20 cm and base 24 cm. Find the radius of its circumcircle (cm).

Shapes inside a sector or quarter circle

r = \frac{R s i n ( θ /2 )}{1 + s i n ( θ /2 )}

A circle inside a sector, a square inside a quarter circle or semicircle — each time draw a right triangle from the centre.

The idea

In these questions a figure (a circle or a square) sits inside a sector of a bigger circle and touches the arc, the radii or the diameter. The same move every time: take the centre $O$ of the big circle, join it to the point of the small figure that lies on the arc, and form a right triangle with $R$ as hypotenuse.

A circle in a sector (angle $θ$ ): the small circle's centre $I$ is on the angle bisector and it touches both radii, so $O I = \frac{r}{s i n ( θ /2 )}$ . It also touches the arc, so $O I + r = R$ , which gives $r = \frac{R s i n ( θ /2 )}{1 + s i n ( θ /2 )}$ . In the figure the circle shrinks into place.

◭ Diagram

Figure: a

6 0^{\circ}

sector; the small circle touches both radii and the arc.

O \to I

2 r

and

2 r + r = R

, so

r = \frac{R}{3}

$θ$	$sin \frac{θ}{2}$	inside circle $r$
$6 0^{\circ}$	$\frac{1}{2}$	$\frac{R}{3}$
$9 0^{\circ}$	$\frac{1}{2}$	$(2 - 1) R$
$12 0^{\circ}$	$\frac{3}{2}$	$(23 - 3) R$
$18 0^{\circ}$	$1$	$\frac{R}{2}$

Letters in the table: $θ$ = angle of the sector ( $9 0^{\circ}$ = quarter circle, $18 0^{\circ}$ = semicircle); $R$ = radius of the big circle; $r$ = radius of the inside circle.

◭ Diagram

Figure: a square (saffron, side

x

) in a quarter circle. Two vertices on the radii at distance

a

from the centre, two on the arc. The blue line is

R

and the right triangle gives

R^{2} = (2 a)^{2} + a^{2}

Other casesA square in a semicircle (side

s

on the diameter): the centre is the middle of the square's base, so

R^{2} = s^{2} + (\frac{s}{2})^{2}

and

R = \frac{s 5}{2}

. A square in a quarter circle with one vertex at the centre and the opposite one on the arc: diagonal

= R

, side

= \frac{R}{2}

Mark the centre $O$ of the big circle and find the point of the figure that lies on the arc.
The line from $O$ to that point is $R$ ; make it the hypotenuse of a right triangle (its legs come from the figure's measurements).
Get $R$ (or the unknown measure) by Pythagoras. For a circle in a sector use $r = \frac{R s i n ( θ /2 )}{1 + s i n ( θ /2 )}$ directly.

◭ Diagram

Figure with the given measures.

Example 28 · Basic

Find the radius of the largest circle that fits inside a semicircle of radius 12 cm, touching the diameter and the arc.

Solution, step by step

1For a semicircle $θ = 18 0^{\circ}$ , $sin 9 0^{\circ} = 1$ .
2 $r = \frac{R \times 1}{1 + 1} = \frac{R}{2} = 6$ cm. (Direct view: the centre of the small circle is $r$ above $O$ , and $r + r = R$ — the circle sits on the radius from $O$ to the arc and its diameter is $R$ .)

Answer 6 cm

sin 9 0^{\circ} = 1

, so it is half.

Example 29 · Standard

A circle inside a

6 0^{\circ}

sector (radius 18 cm) touches both radii and the arc. Find its radius and the distance of its centre from the centre of the sector.

Solution, step by step

1 $θ = 6 0^{\circ}$ , $sin 3 0^{\circ} = \frac{1}{2}$ . $O I = \frac{r}{s i n 3 0 ^{\circ}} = 2 r$ .
2Touching the arc gives $O I + r = R$ : $2 r + r = 18$ , so $r = 6$ cm.
3 $O I = 2 r = 12$ cm. Check: $12 + 6 = 18 = R$ ✓.

Answer $r = 6$ cm, $O I = 12$ cm

In a

6 0^{\circ}

sector

r

is always

\frac{R}{3}

Example 30 · Advanced

In a quarter circle a square has two vertices on the two radii at equal distances from the centre and the other two on the arc. The side of the square is 6 cm. Find the radius of the circle.

Solution, step by step

1Take the centre as the origin. Two vertices are at $(a, 0)$ and $(0, a)$ ; the side between them is $a 2 = 6$ , so $a = \frac{6}{2} = 32$ .
2Going outward along the square, the vertex on the arc is $(a, a)$ away from $(a, 0)$ , i.e. at $(2 a, a)$ .
3It lies on the arc, so $R^{2} = (2 a)^{2} + a^{2} = 5 a^{2} = 5 \times 18 = 90$ and $R = 90 = 310 \approx 9.49$ cm.

Answer $310$ cm

Formula

R = x \frac{5}{2}

6 \times 1.58 = 9.49

✓.

Practice — try yourself

BasicTry it yourself first

Inside a sector of radius 30 cm and angle

6 0^{\circ}

a circle touches both bounding radii and the arc. Find its radius (cm).

BasicTry it yourself first

Inside a sector of radius 20 cm and angle

18 0^{\circ}

a circle touches both bounding radii and the arc. Find its radius (cm).

StandardTry it yourself first

One side of a square lies on the diameter of a semicircle and its two upper vertices lie on the arc. If the side of the square is 8 cm, find the radius of the semicircle (cm).

StandardTry it yourself first

A square is inscribed in a quarter circle so that two of its vertices lie on the two bounding radii at equal distances from the centre and the other two on the arc. The side of the square is 4 cm. Find the square of the radius (sq cm).

StandardTry it yourself first

One side of a square lies on the diameter of a semicircle and its two upper vertices lie on the arc. If the side of the square is 12 cm, find the radius of the semicircle (cm).

AdvancedTry it yourself first

StandardTry it yourself first

Inside a quarter circle of radius 10 cm a circle touches both bounding radii and the arc. Find its radius (cm).

StandardTry it yourself first

A square is drawn in a quarter circle of radius 10 cm with one vertex at the centre and the opposite vertex on the arc. Find the side of the square (cm).

StandardTry it yourself first

One side of a square lies on the diameter of a semicircle and its two upper vertices lie on the arc. If the side of the square is 6 cm, find the radius of the semicircle (cm).

StandardTry it yourself first

Summary, mixed practice and revision

All methods in one place

A table of all methods, exam-style mixed questions and revision cards.

Summary: which method when

Clue · formula

All nine methods in one table: the clue in the question, the formula and the method.

Clue in the question	Rule / formula	Method
Length of a tangent from an outside point	$t = d^{2} - r^{2}$	Tangent from a point
Angle between two tangents	$∠ P + ∠ O = 18 0^{\circ}$	Tangent from a point
Chord joining the points of contact	$A B = \frac{2 r t}{d}$ , $OM = \frac{r ^{2}}{d}$	Chord of contact
A third tangent cuts off a triangle	perimeter $= 2 t$ ; $∠ MON = 9 0^{\circ} - \frac{A}{2}$	Third tangent
Tangent + secant / two secants / chords	$P T^{2} = P A \cdot PB$ · $P A \cdot PB = PC \cdot P D$ · $E A \cdot EB = EC \cdot E D$	Product rule
Angle between tangent and chord	$=$ angle in the alternate segment	Alternate segment
Circle touches a quadrilateral inside	$A B + C D = BC + D A$ ; $Area = rs$	Tangential quadrilateral
Incircle / tangent pieces on the sides	$r = \frac{Δ}{s}$ , $x = s - a$ , $r = \frac{p + q - h}{2}$	Incircle
Distance of circum- and incentre / excircle	$R = \frac{ab c}{4Δ}$ , $r_{a} = \frac{Δ}{s - a}$ , $O I^{2} = R (R - 2 r)$	Circumcircle, excircle
Figure in a sector or quarter circle	$r = \frac{R s i n ( θ /2 )}{1 + s i n ( θ /2 )}$ ; $R = \frac{s 5}{2}$	Shapes in a sector

Letters in the table: $t$ = tangent, $d$ = distance, $r, R$ = radii, $s$ = semi-perimeter, $Δ$ = area, $p, q, h$ = legs and hypotenuse of a right triangle, $θ$ = angle of the sector — each explained in its own method.

OrderOrder in the exam: (1) draw the figure, (2) join the point of contact to the centre and spot the right angle, (3) pick the rule from the table above, (4) check the answer by another route.

Mixed practice

All methods mixed

Exam-style questions from every method with new numbers; first decide which method applies.

In each question first decide: is it a tangent from a point, a product rule, the alternate segment or a circle inside a shape? Then solve, and open the solution to check your route.

BasicTry it yourself first

A circular pond has radius 20 m. A person stands 29 m from its centre and walks straight along a tangent line to the point where it touches the bank. How many metres does the person walk?

StandardTry it yourself first

A circle has radius 24 cm and the tangent drawn from an outside point

P

is 32 cm long. Find the distance of

P

from the centre (cm).

StandardTry it yourself first

P A

and

PB

are tangents to a circle (

A, B

points of contact,

O

the centre) and

∠ A PB = 12 6^{\circ}

. Find

∠ A OB

StandardTry it yourself first

A tangent from a point 41 cm from the centre of a circle is 40 cm long. Find the diameter of the circle (cm).

StandardTry it yourself first

A circle has radius 8 cm and tangents

P A

PB

are drawn from a point

P

at 17 cm from the centre. Find the perimeter of quadrilateral

O A PB

(cm).

StandardTry it yourself first

A circle has radius 30 cm. Tangents

P A

PB

are drawn from a point

P

at 50 cm from the centre. Find the length of the chord of contact

A B

(cm).

StandardTry it yourself first

Tangents are drawn to a circle of radius 15 cm (centre

O

) from an outside point

P

with

OP = 25

cm. Find the distance of the chord of contact from the centre (cm).

AdvancedTry it yourself first

Tangents

P A = PB = 12

cm are drawn to a circle of radius 9 cm from a point

P

. Find the area of triangle

P A B

(sq cm).

StandardTry it yourself first

From an outside point

A

tangents

A B = A C = 25

cm are drawn to a circle. A third tangent cuts

A B

M

and

A C

N

, with

A M = 11

cm and

A N = 12

cm. Find

MN

(cm).

AdvancedTry it yourself first

A B

A C

are tangents to a circle (centre

O

) and

∠ A = 5 8^{\circ}

. A third tangent

MN

cuts

A B

and

A C

M

N

(the circle lies outside triangle

A MN

). Find

∠ MON

StandardTry it yourself first

A circle has radius 8 cm and the outside point

A

is 17 cm from the centre. The tangent at

P

, the point of the circle nearest to

A

, cuts the two tangents from

A

M

and

N

. Find

MN

(cm).

StandardTry it yourself first

From a point

P

a tangent

PT

and a secant

P A B

are drawn to a circle (

A

is the nearer cutting point). If

P A = 8

cm and

A B = 10

cm, find

PT

(cm).

StandardTry it yourself first

Chords

A B

and

C D

of a circle meet inside at

E

. If

A E = 10

EB = 12

and

CE = 15

(cm), find the length of chord

C D

(cm).

AdvancedTry it yourself first

A secant from a point

P

, 25 cm from the centre of a circle of radius 15 cm, cuts the circle at

A

(near) and

B

(far). If

P A = 8

cm, find

PB

(cm).

StandardTry it yourself first

In triangle

A BC

A B = 14

cm and

A C = 21

cm. A circle through

B

and

C

cuts

A B

D

and

A C

E

. If

A D = 6

cm, find

A E

(cm).

AdvancedTry it yourself first

Two circles intersect at

A

and

B

. The line

B A

extended meets their common tangent

PQ

M

. If

M A = 8

cm and

A B = 10

cm, find the tangent segment

PQ

(cm).

AdvancedTry it yourself first

A B

is a diameter of a circle with tangents at

A

and

B

. A third tangent at a point

C

of the arc cuts them at

D

and

E

with

A D = 4

cm and

BE = 9

cm. Find the radius (cm).

AdvancedTry it yourself first

A tangent

PT = 24

cm is drawn from an outside point

P

and a secant from

P

cuts the circle at

A

B

with

P A = 18

cm. If

O

is the centre and

OP = 25

cm, find the radius and the chord

A B

(cm) and give their sum.

StandardTry it yourself first

PQ

is the tangent at

A

to a circle,

A B

is a chord and

∠ P A B = 2 9^{\circ}

. Find

∠ A OB

at the centre

O

StandardTry it yourself first

Triangle

A BC

is inscribed in a circle. The tangent

S A T

A

makes

∠ S A B = 4 4^{\circ}

and

∠ T A C = 5 3^{\circ}

. Find

∠ B A C

AdvancedTry it yourself first

The tangent at

A

to the circumcircle of triangle

A BC

is parallel to

BC

and

∠ B A C = 5 2^{\circ}

. Find

∠ A BC

AdvancedTry it yourself first

A, B, C

are points on a circle. The tangent at

A

meets the chord

BC

produced at

T

(

C

lies between

B

and

T

). If

∠ A TC = 2 8^{\circ}

and

∠ C A T = 3 7^{\circ}

, find the central angle

∠ BOC

StandardTry it yourself first

A circle inside quadrilateral

A BC D

touches all four sides. If

A B = 24

BC = 17

and

C D = 20

cm, find

A D

(cm).

AdvancedTry it yourself first

A trapezium has parallel sides 32 cm and 8 cm, its non-parallel sides are equal and a circle inside touches all four sides. Find the area of the trapezium (sq cm).

AdvancedTry it yourself first

In right trapezium

A BC D

A B ⊥ A D

A B ⊥ BC

A D = 20

cm and

BC = 30

cm. A circle inside touches all four sides. Find the radius (cm).

StandardTry it yourself first

The sides of a triangle are 25, 25 and 14 cm. Find the radius of its incircle (cm).

AdvancedTry it yourself first

The sides forming the right angle of a right triangle are 28 cm and 45 cm. Find the inradius (cm).

StandardTry it yourself first

The incircle of triangle

A BC

touches

BC

C A

A B

D

E

F

. If

A F = 5

B D = 7

and

CE = 8

cm, find the perimeter (cm).

AdvancedTry it yourself first

In triangle

A BC

∠ A = 3 8^{\circ}

and

I

is its incentre. Find

∠ B I C

StandardTry it yourself first

An isosceles triangle has equal sides 13 cm and base 24 cm. Find the radius of its circumcircle (cm).

AdvancedTry it yourself first

The sides of a triangle are 13, 13, 24 cm. Find the distance

O I

(cm) between its circumcentre

O

and incentre

I

AdvancedTry it yourself first

A right triangle has sides 9, 12 and 15 cm. Find the distance (cm) between its circumcentre and incentre.

AdvancedTry it yourself first

In triangle

A BC

BC = 17

C A = 25

A B = 28

cm. Find the radius (cm) of the excircle opposite

C

(touching side

A B

StandardTry it yourself first

Inside a sector of radius 45 cm and angle

6 0^{\circ}

a circle touches both bounding radii and the arc. Find its radius (cm).

AdvancedTry it yourself first

One side of a square lies on the diameter of a semicircle and its two upper vertices lie on the arc. If the side of the square is 20 cm, find the radius of the semicircle (cm).

Revision cards

Tap to flip

A one-line gist of every method: think of the answer first, then flip the card.

KeysLast advice: when stuck, draw the radius to every point of contact in the figure. A tangent gives a right angle, two tangents from a point give equals, a secant gives a product, and a tangent-chord angle gives the alternate segment — those are the four keys.

Tangents and Circle Problems

Why is it 90∘?

What is in this chapter

The idea

Practice — try yourself

The idea

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The idea

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The idea

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The idea

Why is it true?

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The idea

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The idea

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The idea

Practice — try yourself

The idea

Practice — try yourself

Why is it $9 0^{\circ}$ ?