Circles: Chords and Angles

First the names (they appear in every later question), then arc length and sector area from the angle.

Get the words right before the theorems. A radius runs from the centre to a point of the circle. A chord joins two points of the circle. A diameter is a chord that passes through the centre — it is the longest chord and twice the radius. A secant is the whole straight line that cuts the circle at two points (the chord is its part inside), and a tangent touches the circle at one point only (next chapter).

A chord cuts the circle into two arcs: the smaller (minor) arc and the bigger (major) arc; a diameter cuts them equal, and then each arc is a semicircle. The centre joined to the two ends of an arc makes a sector (a slice of pizza); a chord with its arc makes a segment. Two circles with the same centre are concentric.

Now the main idea: the longer the arc, the bigger the angle at the centre. A full turn is 360° and the full circumference is $2\pi r$. So an angle $\theta °$ is the fraction $\frac{\theta }{360}$ of the full turn — the arc is that fraction of the circumference and the sector is that fraction of the area (a quarter of a pizza is a quarter of the rim and a quarter of the area).

$A=\frac{1}{2}\ell r$ is worth remembering: a sector is nearly a “triangle” with base the arc $\ell$ and height $r$. When $\ell$ is given you get the area without finding $\theta$. The area of a segment = sector − triangle (the triangle made by the two radii and the chord). Take $\pi =\frac{22}{7}$ when the radius is a multiple of 7.

1. Decide whether a length (arc) or an area (sector) is asked.
2. Write the angle in degrees (clock: 1 minute = 6°, 1 hour = 30°).
3. Simplify the fraction $\frac{\theta }{360}$, then multiply it by the circumference or the area.
4. Check the unit: length = cm, area = sq cm.

The most useful theorem: the perpendicular from the centre halves the chord; from it, chord, radius and distance questions.

Join the ends of a chord AB to the centre O. $OA$ and $OB$ are both radii, so triangle $OAB$ is isosceles. In an isosceles triangle the perpendicular from the apex to the base cuts the base in half. That is the theorem: the perpendicular from the centre to a chord bisects the chord.

The reverse statements are true too: (1) the line from the centre to the midpoint of a chord is perpendicular to it. (2) The perpendicular bisector of any chord passes through the centre — that is how the centre of a circle is found: draw two chords, and the point where their perpendicular bisectors meet is the centre.

Look at the right triangle $OMB$ it creates: hypotenuse = radius $r$, one leg = distance from the centre $d$, the other leg = half the chord $\frac{c}{2}$. So Pythagoras finds any one of the three when two are given.

Letters: half-chord, $d$ and $r$ are the sides of the right triangle ($r$ the biggest). If you spot the triple you save the square root: half-chord 24 and $r=25$ immediately give $d=7$.

1. Mark the centre $O$; draw the perpendicular from $O$ to the chord and call its foot $M$.
2. $M$ is the midpoint of the chord — write the half-chord $\frac{c}{2}$.
3. Pythagoras in the right triangle $OMB$: $r^2=d^2+(\frac{c}{2}{)}^2$.
4. Find the asked quantity; if the chord is asked, double at the end.

Equal chords are equally far from the centre, make equal angles at the centre and cut equal arcs; the special chords r, r√2, r√3.

Take two equal chords $AB=CD$ and join all four ends to the centre $O$. Triangles $OAB$ and $OCD$ have all three sides equal ($r$, $r$ and the chord), so they are congruent (SSS), and corresponding parts of congruent triangles are equal. Three facts follow:

• Equal angles at the centre: $\angle AOB=\angle COD$ (and equal angles ⟹ equal arcs).
• Equal distances from the centre: the heights ($d$) of the two triangles are equal.
• The reverse is true too: chords equally far from the centre are equal; the chord nearer the centre is longer (and makes the bigger angle).

When the chord and the radius are simply related, the angle at the centre is simple too. The three chords below turn up in every question; in each, $OAB$ is an isosceles triangle with $OA=OB=r$.

Letters: $r$ = radius of the circle. The “distance” column is the same $d$ as before and comes from $d=\sqrt{r^2-(\text{half-chord}{)}^2}$.

1. If the chords are equal, write equal distances from the centre and equal central angles.
2. Use Pythagoras (previous method) to find the missing quantity of one of them.
3. If a chord is $r$, $r\sqrt{2}$ or $r\sqrt{3}$, take the angle straight as 60°, 90°, 120°.

The distance between two parallel chords, or the radius/length from it — with the two cases.

The perpendiculars from the centre to two parallel chords lie on one straight line (both are perpendicular to the chords), so the two distances from the centre are measured along the same line. All that matters is whether the chords are on the same side of the centre or on opposite sides: on the same side the distances subtract, on opposite sides they add.

1. Write the half-length of each chord and find $d=\sqrt{r^2-(\frac{c}{2}{)}^2}$ for each.
2. Read: same side (subtract) or opposite sides (add); if not stated, do both.
3. If the radius is asked (the gap is given), call $d_1=x$, $d_2=x\pm \text{gap}$ and equate the two forms of $r^2$.

From the isosceles triangle formed by two equal chords from one point, find the radius or the third chord.

From a point $A$ of a circle draw two equal chords $AB=AC=a$ and join $BC$. This is an isosceles triangle inside a circle. By symmetry the line $AO$ is perpendicular to the base $BC$ and cuts it at its midpoint $D$ (the isosceles-triangle rule and the perpendicular-from-the-centre rule give the same line).

Let $AD=x$. In triangle $ABD$, $B{D}^2=a^2-x^2$. In triangle $OBD$ the hypotenuse is $OB=r$ and $OD=x-r$ (or $r-x$, same square), so $B{D}^2=r^2-(x-r{)}^2=2rx-x^2$. Equate the two $B{D}^2$: $a^2-x^2=2rx-x^2$ — the $x^2$ cancels and what remains is $a^2=2rx$.

The same $r=\frac{a^2}{2x}$ is the circumradius of any isosceles triangle (equal side $a$, height $x$). When $x<r$ the centre lies beyond $BC$ (the apex angle is obtuse); when $x>r$ the centre is inside the triangle.

1. Take the midpoint $D$ of $BC$; $AD$ passes through the centre and is perpendicular to $BC$.
2. Find $x=AD=\sqrt{a^2-(\frac{BC}{2}{)}^2}$ (Pythagoras).
3. For $r$ use $r=\frac{a^2}{2x}$; for $BC$ first $x=\frac{a^2}{2r}$, then $BC=2\sqrt{a^2-x^2}$.

The length of the shared chord of two intersecting circles and the distance between the centres.

When two circles cut at two points $A$ and $B$, $AB$ is a chord of both — the common chord. Each centre is equally far from $A$ and $B$ (its own radius), so both centres lie on the perpendicular bisector of $AB$. Result: the line of centres $O_1{O}_2$ cuts the common chord at right angles and in half.

Now there are two right triangles $O_{1}MA$ and $O_{2}MA$ sharing $MA=h$. Let $O_{1}M=x$; then $O_{2}M=d-x$ and $h^2=R^2-x^2=r^2-(d-x{)}^2$. Expanding, $x^2$ cancels and $x=\frac{d^2+R^2-r^2}{2d}$. The same formula works when both centres lie on the same side of the chord (then $x>d$).

Special case: if $d^2=R^2+r^2$ (a right angle at $A$ in triangle $O_{1}A{O}_2$; the circles cut at right angles) then, from the altitude formula, $h=\frac{Rr}{d}$ and the common chord is $\frac{2Rr}{d}$. For radii 15 and 20 with centres 25 apart: chord $=\frac{2\times 15\times 20}{25}=24$.

1. Draw both circles and the chord $AB$; mark $M$ where $O_1{O}_2$ meets $AB$.
2. Write the two right triangles $O_{1}MA$ and $O_{2}MA$; $MA=h$ is shared.
3. To find the chord: $x=\frac{d^2+R^2-r^2}{2d}$, then $h=\sqrt{R^2-x^2}$, chord $=2h$.
4. To find $d$: $x_1=\sqrt{R^2-h^2}$, $x_2=\sqrt{r^2-h^2}$; $d=x_1+x_2$ or $|x_1-x_2|$.

The product of the pieces; why it is the same for every chord; the shortest chord.

Through a point $P$ inside a circle draw any chord $AB$; $P$ splits it into pieces $PA$ and $PB$. The surprising fact: the product $PA\times PB$ is the same for every chord through $P$. That is why for two chords crossing at $P$, $PA\times PB=PC\times PD$.

Why? Drop the perpendicular $OM$ from the centre to the chord $AB$ ($M$ its midpoint, half-chord $h$, $h^2=r^2-O{M}^2$). If $P$ is a distance $e$ from $M$, then $PA=h-e$ and $PB=h+e$, so $PA\cdot PB=h^2-e^2=r^2-O{M}^2-e^2$. But in the right triangle $OMP$, $O{M}^2+e^2=O{P}^2$. Hence $PA\cdot PB=r^2-O{P}^2$ — the direction of the chord never appeared, only the distance of $P$ from the centre.

The shortest chord: the farther a chord through $P$ is from the centre, the shorter it is. The farthest is the one perpendicular to $OP$ (distance $=OP$). So the shortest chord is $2\sqrt{r^2-O{P}^2}$ and the longest is the diameter $2r$. If the pieces are equal ($P$ is the midpoint), the product is $P{A}^2$.

1. Draw the figure and write the lengths on the four pieces (the unknown as $x$).
2. Product of the two pieces of one chord = product of the two pieces of the other.
3. Solve; if two roots come, they are the two pieces (like $x$ and $L-x$).
4. For the radius use a chord's product in $=r^2-O{P}^2$.

Two secants from an outside point: whole × outside part; the same rule as inside with the sign reversed.

Now $P$ is outside the circle and a secant from it cuts the circle first at $A$ and then at $B$. The same argument works again: take the midpoint $M$ and half-chord $h$; if $PM=p$ then $PA=p-h$ and $PB=p+h$, so $PA\cdot PB=p^2-h^2=(O{P}^2-O{M}^2)-(r^2-O{M}^2)=O{P}^2-r^2$. That is one number for every secant, so for two secants $PA\times PB=PC\times PD$.

A secant through the centre: if the centre is at distance $d$ from $P$, the near point is at $d-r$ and the far one at $d+r$, so the product is $(d-r)(d+r)=d^2-r^2$ — the same formula. Also far − near $=2r$ (the diameter). When the secant just touches the circle (the two points merge) it is a tangent: $P{T}^2=O{P}^2-r^2$ — the main formula of the next chapter.

1. Draw the figure; for each secant write the “outside part” $PA$ and the “whole” $PB=PA+AB$.
2. Equate the products of the two secants, or put one into $=O{P}^2-r^2$.
3. If an unknown $x$ gives a quadratic take the positive root; if the chord is asked, finish with whole − outside part.