Relative positions of two circles

Two phone masts $O$ and $O'$ are $7$ km apart and cover radii of $5$ km and $3$ km respectively. Do the two coverage zones overlap? What if mast $O'$ is moved so that the masts are $8$ km apart?

Common points of two circles

Explore: pull two circles apart

Take $(O, R)$ and $(O', r)$ with $R\geq r$ , and $d=OO'$ . Put the small circle inside the large one and slowly pull it out: they touch from inside when $d=R-r$ , then intersect, touch from outside when $d=R+r$ , and finally move apart.

The line $OO'$ is called the line of centres. Every line through the centre is a line of symmetry of a circle, so the line of centres is a common line of symmetry of both circles.

Two intersecting circles

Why? We have $OA=OB$ and $O'A=O'B$ , so $O$ and $O'$ both lie on the perpendicular bisector of $AB$ .

If two circles touch at $T$ and $T$ is not on $OO'$ , then the reflection of $T$ in $OO'$ is also a common point (because $OO'$ is a common line of symmetry). Then the circles would have two common points, which is impossible.

Two circles intersecting at

A

and

B

Example The length of a common chord

The circles $(O, 15\text{ cm})$ and $(O', 13\text{ cm})$ intersect at $A$ and $B$ , with $OO'=14$ cm. Find $AB$ .

Solution. Let $H$ be the point where $OO'$ meets $AB$ ; then $AH\perp OO'$ and $H$ is the midpoint of $AB$ . Let $OH=x$ ; then $O'H=14-x$ (because $H$ lies between $O$ and $O'$ ). By Pythagoras' theorem in the right-angled triangles $AHO$ and $AHO'$ , we have $AH^2=15^2-x^2=13^2-(14-x)^2$ , hence

$225-x^2=169-196+28x-x^2\Rightarrow 28x=252\Rightarrow x=9.$

Therefore $AH=\sqrt{15^2-9^2}=12$ (cm) and $AB=2AH=24$ cm.

So $AB=24$ cm.

Two touching circles

When two circles touch externally at $T$ , the point $T$ lies between $O$ and $O'$ , so $OO'=OT+TO'=R+r$ . The line through $T$ perpendicular to $OO'$ is a common tangent of the two circles at $T$ .

Two circles touching externally at

T

Key terms

English	Tiếng Việt
line of centres	đường nối tâm
common chord	dây chung
touch externally	tiếp xúc ngoài
touch internally	tiếp xúc trong
lie outside each other	ở ngoài nhau
one inside the other	đựng nhau
common tangent	tiếp tuyến chung

Summary

Intersect: $R-r<d<R+r$ ; touch externally: $d=R+r$ ; touch internally: $d=R-r$ ; outside each other: $d>R+r$ ; one inside the other: $d<R-r$ .
The line of centres is a line of symmetry, the perpendicular bisector of the common chord, and passes through the point of contact.
For a common chord: let $OH=x$ and write $AH^2$ in two ways.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Consider the circles $(O, 6\text{ cm})$ and $(O', 4\text{ cm})$ . Find their relative position when $OO'$ is: a) $11$ cm; b) $10$ cm; c) $5$ cm; d) $2$ cm; e) $1$ cm.

What you need and how to start

What you need: compare $d$ with $R+r=10$ and $R-r=2$ .

Steps:

Compute $R+r$ and $R-r$ , then compare each value with them.

Show solution

We have $R+r=10$ and $R-r=2$ . Hence: a) $11>10$ ; b) $10=R+r$ ; c) $2<5<10$ ; d) $2=R-r$ ; e) $1<2$ .

So a) outside each other; b) touch externally; c) intersect; d) touch internally; e) one inside the other.

Answer. a) outside each other; b) touch externally; c) intersect; d) touch internally; e) one inside the other.

Exercise 2

The circles $(O, 5\text{ cm})$ and $(O', 5\text{ cm})$ intersect at $A$ and $B$ , with $OO'=8$ cm. Find $AB$ .

What you need and how to start

What you need: $OO'$ is the perpendicular bisector of $AB$ ; $H$ is the midpoint of $OO'$ .

Steps:

Find $OH=4$ cm, then $AH$ by Pythagoras' theorem.

Show solution

Let $H$ be the point where $OO'$ meets $AB$ . The two circles have equal radii, so $H$ is the midpoint of $OO'$ , hence $OH=4$ cm. In the triangle $AHO$ , right-angled at $H$ , we have $AH=\sqrt{5^2-4^2}=3$ (cm), so $AB=6$ cm.

So $AB=6$ cm.

Answer. $AB=6$ cm.

Exercise 3

The circle $(O, 9\text{ cm})$ touches the circle $(O', r)$ internally, with $r<9$ cm and $OO'=4$ cm. Find $r$ .

What you need and how to start

What you need: touching internally means $d=R-r$ .

Steps:

Solve the equation $9-r=4$ .

Show solution

The circles touch internally, so $9-r=4$ , hence $r=5$ (cm).

So $r=5$ cm.

Answer. $r=5$ cm.

Exercise 4

The circles $(O, 7\text{ cm})$ and $(O', 3\text{ cm})$ touch each other. Find $OO'$ .

What you need and how to start

What you need: touching externally: $d=R+r$ ; touching internally: $d=R-r$ .

Steps:

Look at the two cases.

Show solution

If they touch externally, $OO'=7+3=10$ (cm). If they touch internally, $OO'=7-3=4$ (cm).

So $OO'=10$ cm or $OO'=4$ cm.

Answer. $OO'=10$ cm or $OO'=4$ cm.

Exercise 5

A logo is made of two circles of radii $3$ cm and $4$ cm whose centres are $5$ cm apart. Minh needs to draw the common chord of the two circles. Find the length of the common chord. (Revise equations reducible to linear equations.)

What you need and how to start

What you need: let $OH=x$ and write $AH^2$ in two ways.

Steps:

Form the equation $3^2-x^2=4^2-(5-x)^2$ .
Find $AH$ , then $AB=2AH$ .

Show solution

Let $O$ , $O'$ be the centres of the circles of radii $3$ cm and $4$ cm; let $H$ be the point where $OO'$ meets the common chord $AB$ , and $OH=x$ . We have

$9-x^2=16-(5-x)^2\Rightarrow 9=-9+10x\Rightarrow x=1.8.$

Hence $AH=\sqrt{9-1.8^2}=\sqrt{5.76}=2.4$ (cm), so $AB=4.8$ cm.

So the common chord is $4.8$ cm long.

Answer. $4.8$ cm.

Exercise 6

Three circles with centres $A$ , $B$ , $C$ touch one another externally in pairs, with $AB=7$ cm, $BC=8$ cm and $CA=9$ cm. Find the radius of each circle. (Revise solving simultaneous equations.)

What you need and how to start

What you need: when two circles touch externally, the distance between the centres is the sum of the radii.

Steps:

Form three equations in the unknowns $a$ , $b$ , $c$ , then add them.

Show solution

Let $a$ , $b$ , $c$ (cm) be the radii of the circles with centres $A$ , $B$ , $C$ ( $a$ , $b$ , $c>0$ ). From the question, $a+b=7$ , $b+c=8$ and $c+a=9$ . Adding the three equations gives $2(a+b+c)=24$ , so $a+b+c=12$ . Hence

$c=12-7=5,\qquad a=12-8=4,\qquad b=12-9=3\ \text{(satisfying the conditions)}.$

So the circles with centres $A$ , $B$ , $C$ have radii $4$ cm, $3$ cm and $5$ cm respectively.

Answer. $4$ cm, $3$ cm, $5$ cm (centres $A$ , $B$ , $C$ ).

Exercise 7

(1.5 marks) The circles $(O, 4\text{ cm})$ and $(O', 2\text{ cm})$ touch externally at $A$ . The line $a$ passes through $A$ and is perpendicular to $OO'$ .

a) Find $OO'$ .

b) Prove that $a$ is a common tangent of the two circles.

c) Take a point $M$ on $a$ with $MA=3$ cm. Find $MO$ and $MO'$ (give your answers to 2 decimal places).

What you need and how to start

What you need: the point of contact lies on the line of centres; the test for a tangent.

Steps:

Use $OO'=R+r$ .
Show that $a\perp OA$ and $a\perp O'A$ at $A$ .
Use Pythagoras' theorem in the triangles right-angled at $A$ .

Show solution

a) The circles touch externally, so $OO'=4+2=6$ (cm).

b) The point $A$ lies on $OO'$ , so $a\perp OA$ and $a\perp O'A$ at $A$ . By the test for a tangent, $a$ is the tangent to both $(O)$ and $(O')$ at $A$ .

c) In the triangles $MAO$ and $MAO'$ , right-angled at $A$ , we have

$MO=\sqrt{3^2+4^2}=5\ \text{(cm)},\qquad MO'=\sqrt{3^2+2^2}=\sqrt{13}\approx3.61\ \text{(cm)}.$

So $OO'=6$ cm, $MO=5$ cm and $MO'\approx3.61$ cm.

Answer. a) $OO'=6$ cm; b) see the proof; c) $MO=5$ cm, $MO'=\sqrt{13}\approx3.61$ cm.

Exercise 8 Challenge

The circles $(O, 9\text{ cm})$ and $(O', 4\text{ cm})$ touch externally. A common tangent touches $(O)$ at $B$ and $(O')$ at $C$ , and does not pass through the point where the circles touch. Find $BC$ .

What you need and how to start

What you need: $OB\perp BC$ and $O'C\perp BC$ (property of a tangent).

Steps:

Draw $O'K\perp OB$ at $K$ ; show that $BCO'K$ is a rectangle.
Find $OK$ and $OO'$ , then $O'K$ by Pythagoras' theorem.

Show solution

We have $OB\perp BC$ and $O'C\perp BC$ , so $OB\parallel O'C$ . Draw $O'K\perp OB$ at $K$ ; the quadrilateral $BCO'K$ has three right angles, so it is a rectangle. Hence $BC=O'K$ and $BK=O'C=4$ cm, so $OK=9-4=5$ cm. The circles touch externally, so $OO'=9+4=13$ cm. In the triangle $OKO'$ , right-angled at $K$ , we have

$O'K=\sqrt{13^2-5^2}=12\ \text{(cm)}.$

So $BC=12$ cm.

Answer. $BC=12$ cm.

a) The two circles intersect.	True	False
b) The line of centres $OO'$ is perpendicular to the common chord.	True	False
c) The two circles touch internally.	True	False
d) The common chord has length $6$ cm.	True	False

Common points of two circles

Explore: pull two circles apart

Two intersecting circles

Two touching circles

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer