Relative positions of a line and a circle

A mobile phone mast covers a radius of $5$ km. One straight main road is $3$ km from the mast, and another road is $7$ km from it. How long is the stretch with signal on each road?

Three positions of a line and a circle

Explore: move the line away from the centre

Let $d=OH$ be the distance from $O$ to the line $a$ ( $OH\perp a$ ). Move $a$ slowly away from the centre: at first there are two points of intersection, they merge into one when $d=R$ , and then they disappear when $d>R$ .

When $a$ touches $(O)$ , $a$ is called a tangent and the common point is the point of contact. When they intersect, the two points form a chord of length $2\sqrt{R^2-d^2}$ (lesson Arcs and chords).

The line

a

cuts,

b

touches and

c

does not meet

(O)

Tangents to a circle

Indeed, $a\perp OA$ at $A$ exactly when the distance from $O$ to $a$ is $d=OA=R$ .

Because of the right angle at the point of contact, we find the length of a tangent by Pythagoras' theorem.

Two tangents from a point

Why? The triangles $OAM$ and $OBM$ are right-angled at $A$ and $B$ , share the hypotenuse $OM$ and have $OA=OB$ , so they are congruent (RHS). Hence $MA=MB$ , $\angle AMO=\angle BMO$ and $\angle AOM=\angle BOM$ .

Two tangents

MA

MB

to the circle

(O)

Example Two tangents from one point

From a point $M$ with $OM=10$ cm, draw the two tangents $MA$ , $MB$ to the circle $(O, 5\text{ cm})$ . Find $MA$ , the angle $\angle AMB$ and $AB$ .

Solution. In the triangle $OAM$ , right-angled at $A$ , we have

$MA=\sqrt{10^2-5^2}=5\sqrt3\approx8.66\ \text{(cm)},\qquad \sin\angle AMO=\frac{OA}{OM}=\frac12\Rightarrow\angle AMO=30^\circ.$

Since $MO$ bisects $\angle AMB$ , we get $\angle AMB=60^\circ$ . The triangle $MAB$ is isosceles with an angle of $60^\circ$ , so it is equilateral, hence $AB=MA$ .

So $MA=AB=5\sqrt3\approx8.66$ cm and $\angle AMB=60^\circ$ .

Key terms

Summary

$d<R$ : they intersect (chord of length $2\sqrt{R^2-d^2}$ ); $d=R$ : they touch; $d>R$ : they do not meet.
A tangent is perpendicular to the radius at the point of contact, and conversely.
Length of a tangent: $MA=\sqrt{OM^2-R^2}$ .
Two tangents from $M$ : $MA=MB$ , and $MO$ , $OM$ are angle bisectors.

Practice

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

Exercise 1

Three lines are at distances $4$ cm, $6$ cm and $7.5$ cm from the centre of the circle $(O, 6\text{ cm})$ . Find the position of each line relative to the circle.

What you need and how to start

What you need: compare $d$ with $R$ .

Steps:

Compare each distance with $6$ .

Show solution

We have $4<6$ , $6=6$ and $7.5>6$ .

So the first line cuts the circle, the second touches it and the third does not meet it.

Answer. The first line cuts the circle, the second touches it, the third does not meet it.

Exercise 2

The line $a$ is at distance $8$ cm from the centre of the circle $(O, 10\text{ cm})$ . Prove that $a$ cuts the circle at two points $A$ , $B$ and find $AB$ .

What you need and how to start

What you need: if $d<R$ they intersect; the chord has length $2\sqrt{R^2-d^2}$ .

Steps:

Compare $8$ with $10$ , then find the chord.

Show solution

We have $d=8<10=R$ , so $a$ cuts the circle at two points. Then $AB=2\sqrt{10^2-8^2}=2\cdot6=12$ (cm).

So $AB=12$ cm.

Answer. $AB=12$ cm.

Exercise 3

From a point $M$ at distance $15$ cm from the centre $O$ , draw the tangent $MA$ to the circle $(O, 9\text{ cm})$ . Find $MA$ .

What you need and how to start

What you need: $OA\perp MA$ , so the triangle $OAM$ is right-angled at $A$ .

Steps:

Use Pythagoras' theorem with hypotenuse $OM$ .

Show solution

In the triangle $OAM$ , right-angled at $A$ , we have $MA=\sqrt{15^2-9^2}=\sqrt{144}=12$ (cm).

So $MA=12$ cm.

Answer. $MA=12$ cm.

Exercise 4

The triangle $ABC$ is right-angled at $A$ , with $AB=3$ cm and $AC=4$ cm. Prove that $AB$ is a tangent to the circle $(C, CA)$ and that the point $B$ lies outside this circle.

What you need and how to start

What you need: the test for a tangent.

Steps:

Show that $A$ is on $(C, CA)$ and $AB\perp CA$ .
Compare $CB$ with $4$ .

Show solution

We have $A$ on $(C, CA)$ and $AB\perp CA$ at $A$ , so $AB$ is a tangent. By Pythagoras' theorem, $CB=\sqrt{3^2+4^2}=5>4$ , so $B$ lies outside the circle.

So $AB$ is a tangent and $B$ lies outside $(C, CA)$ .

Answer. $AB$ is a tangent to $(C, CA)$ ; $B$ lies outside the circle because $CB=5$ cm $>4$ cm.

Exercise 5

From the top $M$ of a tower $80$ m high, we look out to sea to the horizon at $A$ , where $MA$ touches the Earth. Take a cross-section of the Earth to be a circle with centre $O$ and radius $6400$ km. Find $MA$ (give your answer to the nearest kilometre).

What you need and how to start

What you need: $OA\perp MA$ ; convert $80$ m $=0.08$ km.

Steps:

Compute $OM=6400+0.08$ (km).
Use Pythagoras' theorem in the triangle $OAM$ , right-angled at $A$ .

Show solution

We have $OM=6400.08$ km and the triangle $OAM$ is right-angled at $A$ , so

$MA=\sqrt{6400.08^2-6400^2}\approx\sqrt{1024}=32\ \text{(km)}.$

So from the top of the tower we can see about $32$ km.

Answer. About $32$ km.

Exercise 6

From a point $M$ with $OM=8$ cm, draw the two tangents $MA$ , $MB$ to the circle $(O, 4\text{ cm})$ . Find the angle $\angle AMB$ and the measure of the minor arc $AB$ . (Revise arcs and chords.)

What you need and how to start

What you need: $\sin\angle AMO=\dfrac{OA}{OM}$ ; the angles of a quadrilateral add up to $360^\circ$ .

Steps:

Find $\angle AMO$ , then $\angle AMB$ , then $\angle AOB$ .

Show solution

In the triangle $OAM$ , right-angled at $A$ , we have $\sin\angle AMO=\dfrac48=\dfrac12$ , so $\angle AMO=30^\circ$ and $\angle AMB=60^\circ$ . The quadrilateral $OAMB$ has right angles at $A$ and $B$ , so $\angle AOB=360^\circ-90^\circ-90^\circ-60^\circ=120^\circ$ .

So $\angle AMB=60^\circ$ and the minor arc $AB$ measures $120^\circ$ .

Answer. $\angle AMB=60^\circ$ ; the minor arc $AB$ measures $120^\circ$ .

Exercise 7

(2.0 marks) From a point $M$ outside the circle $(O, 3\text{ cm})$ , with $OM=5$ cm, draw the two tangents $MA$ , $MB$ ( $A$ , $B$ are the points of contact). Let $H$ be the point where $OM$ meets $AB$ .

a) Prove that the four points $M$ , $A$ , $O$ , $B$ lie on one circle.

b) Prove that $OM\perp AB$ at $H$ .

c) Find $MA$ and $AB$ .

What you need and how to start

What you need: the median to the hypotenuse is half the hypotenuse; the property of the perpendicular bisector.

Steps:

Let $I$ be the midpoint of $OM$ and prove that $IA=IB=IO=IM$ .
Show that $OM$ is the perpendicular bisector of $AB$ .
Find $AH$ using the area of the triangle $OAM$ .

Show solution

a) The triangles $OAM$ and $OBM$ are right-angled at $A$ and $B$ (property of a tangent). Let $I$ be the midpoint of $OM$ ; then $IA=IB=IO=IM=\dfrac{OM}{2}$ (the median to the hypotenuse). Hence $M$ , $A$ , $O$ , $B$ lie on the circle with centre $I$ .

b) We have $MA=MB$ (two tangents from a point) and $OA=OB$ , so $OM$ is the perpendicular bisector of $AB$ . Hence $OM\perp AB$ at the midpoint $H$ of $AB$ .

c) We have $MA=\sqrt{5^2-3^2}=4$ (cm). The area of the triangle $OAM$ is $\dfrac12\cdot OA\cdot MA=\dfrac12\cdot OM\cdot AH$ , so $AH=\dfrac{3\cdot4}{5}=2.4$ (cm). Hence $AB=2AH=4.8$ cm.

So $MA=4$ cm and $AB=4.8$ cm.

Answer. a), b) See the proof; c) $MA=4$ cm, $AB=4.8$ cm.

Exercise 8 Challenge

From a point $M$ outside the circle $(O)$ , draw the two tangents $MA$ , $MB$ with $MA=10$ cm. The tangent at a point $C$ on the minor arc $AB$ meets $MA$ and $MB$ at $D$ and $E$ respectively. Find the perimeter of the triangle $MDE$ .

What you need and how to start

What you need: the property of two tangents from a point, used at $D$ and at $E$ .

Steps:

Show that $DC=DA$ and $EC=EB$ , then write the perimeter in terms of $MA$ , $MB$ .

Show solution

We have $DC=DA$ (two tangents from $D$ ) and $EC=EB$ (two tangents from $E$ ). Hence

$MD+DE+EM=MD+DC+CE+EM=(MD+DA)+(EB+EM)=MA+MB=20\ \text{(cm)}.$

So the perimeter of the triangle $MDE$ is $20$ cm.

Answer. $20$ cm.

English	Tiếng Việt
intersect	cắt nhau
touch (be tangent)	tiếp xúc
do not meet	không giao nhau
tangent	tiếp tuyến
point of contact	tiếp điểm
angle bisector	tia phân giác

a) $MA=6\sqrt3$ cm.	True	False
b) $\angle AMO=30^\circ$ .	True	False
c) $\angle AMB=30^\circ$ .	True	False
d) $MA=MB$ .	True	False

Three positions of a line and a circle

Explore: move the line away from the centre

Tangents to a circle

Two tangents from a point

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer