Two Circles and Touching Circles

The five positions of two circles and the number of common tangents in each.

The idea

A common tangent is a line touching both circles. If both circles lie on the same side of it, it is a direct common tangent; if they lie on opposite sides, it is a transverse one, passing between the circles.

In the figure the small circle slides towards the big one. The distance $d$ falls and the number of tangents (in the corner circle) goes $4\to 3\to 2\to 1\to 0$. The transverse (saffron) lines last only while the circles are apart ($d\ge R+r$); the direct (pink) lines last until one circle hides completely inside the other ($d\ge R-r$).

Letters in the table: $R$ = radius of the bigger circle, $r$ = of the smaller, $d$ = distance between the two centres. Concentric circles ($d=0$) have no common tangent.

1. Find $R+r$ and $|R-r|$.
2. Compare $d$ with both and pick the row of the table. Equality means touching.
3. If “how many values of $d$” is asked count the whole numbers strictly between the limits.

Practice — try yourself

The length of the direct and the transverse common tangent, $2\sqrt{Rr}$ for touching circles and the tangent at the contact point.

The idea

Two circles: $O_1$ (radius $R$), $O_2$ (radius $r$), centres $d$ apart. A direct common tangent $AB$ ($A,B$ points of contact) is perpendicular to both radii $O_{1}A$, $O_{2}B$, so they are parallel. Drop $O_{2}C$ perpendicular to $O_{1}A$: $AB{O}_{2}C$ is a rectangle ($O_{2}C=AB$, $AC=r$) and in the right triangle $O_{1}C{O}_2$ we have $O_{1}C=R-r$, hypotenuse $O_1{O}_2=d$.

For a transverse tangent the circles lie on opposite sides. Extend $O_{1}A$ beyond $A$ by $r$ to $C$ so that $O_{2}C\parallel AB$; now $O_{1}C=R+r$ and the same kind of right triangle appears.

1. Decide whether the question asks for the direct ($R-r$) or the transverse ($R+r$) tangent; for touching circles direct $=2\sqrt{Rr}$.
2. Make the right triangle: hypotenuse $d$, one side $R\mp r$, the other the tangent.
3. Find the unknown side by Pythagoras (always identify the hypotenuse $d$ separately).

Practice — try yourself

Circles touching externally and internally: distance of the centres, the point of contact and a chord linked to the inside circle.

The idea

Two circles touch when they share exactly one point $T$. Then both centres and $T$ lie on one straight line (the common tangent at $T$ is perpendicular to both radii). In external touching $T$ lies between the centres, so $d=R+r$; in internal touching $T$ is on the same side of both centres, so $d=R-r$.

A chord linked to the inside circle

A small circle (radius $r$) touches the big circle (radius $R$) internally at $T$. The chord of the big circle that touches the small circle on the side opposite to $T$ is perpendicular to the line of centres and is at distance $|R-2r|$ from the big centre (the small centre is $R-r$ away, minus $r$). Half-chord $=\sqrt{R^2-(R-2r{)}^2}=2\sqrt{r(R-r)}$, so the whole chord $=4\sqrt{r(R-r)}$.

1. External touching $\to d=R+r$; internal $\to d=R-r$. The point of contact is on the line of centres.
2. Find the unknown radius from this equation.
3. For a chord question find the distance of the chord from the centre and make a right triangle (hypotenuse $R$).

Practice — try yourself

The third circle between two touching circles and their tangent line, and a chain of circles between two lines.

The third circle

Two circles (radii $a$, $b$) touch externally and touch a line; in the gap below them a third circle (radius $c$) touches both circles and the line. By the previous method the distances along the line between the points of contact are $MN=2\sqrt{ab}$, $MP=2\sqrt{ac}$, $PN=2\sqrt{bc}$. Since $P$ lies between, $MN=MP+PN$.

Circles between two lines

Two lines meet at $V$ and a chain of circles sits on the bisector, all touching both lines. If each line makes an angle $\alpha$ with the bisector, a circle's centre is $\frac{r}{\sin \alpha }$ from $V$. Two successive circles touch, so $\frac{r_2-r_1}{\sin \alpha }=r_1+r_2$ and $\frac{r_2}{r_1}=\frac{1+\sin \alpha }{1-\sin \alpha }=k$ — the radii grow in a geometric progression.

Letters in the table: $\alpha$ = the angle between each line and the bisector (the full angle between the lines is $2\alpha$); $k$ = ratio of the radius of the next circle to the previous.

1. For a third circle use $\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}$ — find $\sqrt{a},\sqrt{b}$ first.
2. For circles between two lines get $k=\frac{1+\sin \alpha }{1-\sin \alpha }$ from $\sin \alpha$ and keep multiplying the radii by $k$.
3. For a distance use $x/(x+L)=r_1/r_2$ to find $x$.

Practice — try yourself

The triangle of the centres of three circles: radii $s-$ the opposite side, right angle and area.

The idea

Three circles (radii $r_1,r_2,r_3$) touch pairwise externally, so the distance between the centres of each pair is the sum of their radii. So the triangle of centres $O_1{O}_2{O}_3$ has sides $r_1+r_2$, $r_2+r_3$, $r_1+r_3$. Its semi-perimeter is $s=r_1+r_2+r_3$ and each radius $=s-$(the side opposite to it). It is exactly the incircle idea ($x=s-a$) — the three circles meet at the points where the incircle of the triangle of centres touches the sides.

1. Draw the triangle of centres: sides = sums of the related radii.
2. If distances are given find $s$ and then $r_1=s-O_2{O}_3$ etc.; if radii are given build the sides by adding.
3. If a right angle or an area is mentioned apply Pythagoras or Heron to that triangle.

Practice — try yourself

Two circles with the same centre: the chord of the outer circle touching the inner one, the ring area and related questions.

The idea

Two circles with the same centre $O$ are concentric; the outer has radius $R$, the inner $r$. A chord of the outer circle touching the inner circle at $D$ is a tangent of the inner circle, so $OD\perp$ the chord. A perpendicular from the centre bisects the chord, so $D$ is the mid-point of the chord.

1. Draw the radius $r$ from $O$ to the point of contact $D$; $OD\perp$ the chord.
2. In the right triangle $ODB$ with hypotenuse $R$: half-chord $=\sqrt{R^2-r^2}$.
3. For a ring or a width use $R^2-r^2=(R+r)(R-r)$.

Practice — try yourself

The bisector of angle $A$ of a triangle meets side $BC$ and the circumcircle: three product rules and the special property of $E$.

The idea

In the circumcircle of triangle $ABC$ the bisector of $\angle A$ meets $BC$ at $D$ and the circle again at $E$. Equal angles ($\angle BAE=\angle EAC$) cut equal arcs, so $E$ is the mid-point of arc $BC$ and $EB=EC$. This one fact gives all three rules.

Letters: $D$ = where the bisector meets $BC$; $E$ = its second meeting point with the circumcircle; $AE=AD+DE$. Also: $EB=EC=EI$ ($I$ the incentre, which lies on $AE$). Angles: $\angle EBC=\angle EAC=\frac{A}{2}$ and $\angle BEC=180^{\circ }-A$. Division of the side: $\frac{BD}{DC}=\frac{AB}{AC}$.

1. In the figure $A$, $D$, $E$ lie on a line; write $AE=AD+DE$.
2. From the three given quantities choose exactly the rule that contains them.
3. Solve for the unknown; if $DE$ or $AD$ is asked subtract from $AE$.

Practice — try yourself

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