Chapter 10 · Circles

10.1 Tangent to a Circle

A straight line and a circle can cut, touch or miss each other. The line that touches a circle at exactly one point is a tangent — and it is always perpendicular to the radius drawn to that point. Learn the proof, and watch it work.

Lay a ruler across a coin. If you slide the ruler away from the coin's centre, it first cuts the edge of the coin at two points, then there is a moment when it just touches the edge, and after that it misses the coin completely. That one special moment, with a single point of contact, is the heart of this section.

We first give names to the three positions of a line and a circle. Then we prove the first of the two big theorems of the chapter: the tangent at a point of a circle is perpendicular to the radius through that point. That right angle is what lets you use Pythagoras' theorem in every tangent problem.

Secants and tangents

Take a circle with centre $O$ and radius $r$, and a straight line $\ell$ in the same plane. There are exactly three possibilities.

📘 Definition 10.1 — Secant

A line that meets a circle at two distinct points is called a secant of the circle. The segment of the secant between the two points is a chord.

📘 Definition 10.2 — Tangent and point of contact

A line that meets a circle at exactly one point is called a tangent to the circle. The single common point is called the point of contact (or the point of tangency), and we say the line touches the circle at that point.

The third possibility is a line with no common point with the circle: it lies completely outside. Think of a secant as a tangent line that has been pushed in a little, so that the two intersection points of a secant move closer and closer together. When they come together and become one point, the secant has turned into a tangent. The first lab shows exactly this.

The circle has radius $r=5$ cm. The vertical line $\ell$ is at perpendicular distance $d$ from the centre $O$ (the blue dashed segment). Slide $d$ or press Play to sweep the line towards the circle. While $d\lt5$ the line is a secant (red, with two points); when $d=5$ it is a tangent (green, with one point); when $d\gt5$ it misses the circle. Watch the length of the chord shrink to zero.

Distance $d$ of the line from $O$ (cm)7.00

distance $d$ and radius $r$ (cm)—

number of common points—

chord length $2\sqrt{r^2-d^2}$ (cm)—

the line is a…—

The tangent is perpendicular to the radius

📐 Theorem 10.1 — Tangent and radius

The tangent at any point of a circle is perpendicular to the radius drawn through the point of contact.

Proof. Let a tangent $XY$ touch the circle with centre $O$ at the point $P$. We must show that $OP\perp XY$.

Step 1. Take any point $Q$ on the tangent other than $P$. The tangent meets the circle only at $P$, so $Q$ is not on the circle. Since the tangent does not enter the circle, $Q$ lies outside it, and therefore $OQ\gt OP$ (because $OP=r$ and points outside the circle are at distance more than $r$ from $O$).
Step 2. So $OP$ is shorter than $OQ$ for every point $Q\ne P$ of the line $XY$. Thus $P$ is the point of $XY$ that is closest to $O$.
Step 3. From Class 9 we know that the shortest distance from a point to a line is along the perpendicular. The closest point of $XY$ to $O$ is $P$, so $OP$ is that perpendicular: $OP\perp XY$. $\blacksquare$

What the theorem gives you. Whenever a tangent is drawn, join the centre to the point of contact. You immediately have a right angle, $\angle OPQ=90^\circ$, at the point of contact. Then:

for a point $T$ on the tangent, $\triangle OPT$ is right-angled at $P$, so $OT^2=OP^2+PT^2=r^2+PT^2$ (Pythagoras), and the other two angles add up to $90^\circ$;
the perpendicular distance from the centre to a tangent is exactly the radius.

The converse is true as well. Suppose a line passes through a point $P$ of the circle and is perpendicular to the radius $OP$. Every other point $Q$ of the line is the far end of the hypotenuse $OQ$ of a right-angled triangle $OPQ$, so $OQ\gt OP=r$: $Q$ is outside the circle. So the line has no point in common with the circle except $P$, and it is a tangent. This is also the reason behind the distance test:

Perpendicular distance $d$ of the line from $O$	Common points	The line is a…
$d\lt r$	two	secant
$d=r$	exactly one	tangent (the foot of the perpendicular is the point of contact)
$d\gt r$	none	line outside the circle

How many tangents? Through a point inside a circle, every line is a secant, so there is no tangent. Through a point on the circle there is exactly one tangent, the line perpendicular to the radius at that point. From a point outside the circle you can draw exactly two tangents; we study them in the next section.

⚠️ Three things to watch

1. Do not forget the right angle. In a tangent figure, the first step is nearly always "join the centre to the point of contact", which gives $\angle OPT=90^\circ$. Missing this is the commonest reason students get stuck.

2. The right angle is at the point of contact, not at $T$. In $\triangle OPT$ the hypotenuse is $OT$ (from the centre to the outside point), not $PT$. So $OT^2=r^2+PT^2$, never $PT^2=r^2+OT^2$.

3. A tangent never passes through the centre. A line through the centre meets the circle at two points (the ends of a diameter), so it is a secant, not a tangent.

🧭 Problem-solving strategy — tangent problems

Mark the centre $O$ and the point of contact $P$, and join them. Mark $\angle OPT=90^\circ$.
Identify the right-angled triangle $OPT$ with the right angle at $P$. The hypotenuse is $OT$.
Use Pythagoras ($OT^2=OP^2+PT^2$) for lengths, or the angle sum ($90^\circ$ for the two acute angles) for angles.
Check that the hypotenuse is the longest side and that all lengths are positive.

✏️ Example 1 — A tangent from a point at a known distance

A circle has centre $O$ and radius $5$ cm. The point $T$ is $13$ cm from $O$, and the tangent from $T$ touches the circle at $P$. Find $TP$ and the area of $\triangle OPT$.

💡 Show solution▶

Right angle $OP\perp TP$ (Theorem 10.1), so $\triangle OPT$ is right-angled at $P$ with hypotenuse $OT=13$ cm.

Pythagoras $TP^2=OT^2-OP^2=13^2-5^2=169-25=144$, so $TP=12$ cm.

Area The legs are $OP$ and $TP$: area $=\tfrac12\times5\times12=30$ cm$^2$.

Check $5^2+12^2=25+144=169=13^2$ ✓.

✏️ Example 2 — Finding an angle

$PQ$ is a tangent to a circle with centre $O$, touching it at $P$. If $\angle POQ=50^\circ$, find $\angle PQO$.

💡 Show solution▶

Right angle $\angle OPQ=90^\circ$, since the tangent $PQ$ is perpendicular to the radius $OP$.

Angle sum In $\triangle OPQ$: $\angle PQO=180^\circ-90^\circ-50^\circ=40^\circ$.

Check $90^\circ+50^\circ+40^\circ=180^\circ$ ✓.

✏️ Example 3 — Two concentric circles

Two concentric circles have radii $5$ cm and $3$ cm. A chord of the larger circle touches the smaller circle. Find the length of the chord.

💡 Show solution▶

Figure Let the chord $AB$ touch the smaller circle at $P$. Then $AB$ is a tangent to the smaller circle, so $OP\perp AB$ with $OP=3$ cm. Also $OA=5$ cm (radius of the larger circle).

Half the chord In right-angled $\triangle OPA$: $AP^2=OA^2-OP^2=25-9=16$, so $AP=4$ cm.

Whole chord The perpendicular from the centre to a chord bisects the chord (Class 9), so $AB=2\times AP=8$ cm.

Check $3^2+4^2=5^2$ ✓.

The circle has radius $r=3$ cm. Choose the point of contact $P$ with the first slider (or press Play). The tangent at $P$ is the red line. Slide the point $Q$ along the tangent away from $P$ with the second slider: every such $Q$ is farther from $O$ than $r$ (the green dashed segment $OQ$ is longer than the radius). So $P$ is the closest point of the tangent to $O$, and the angle at $P$ is $90^\circ$.

Point of contact $P$ (angle)30°Distance $PQ$ along the tangent (cm)4.0

$\angle OPQ$—

$OQ$ (cm) and $r$ (cm)—

$OQ^2$—

$r^2+PQ^2$—

Using the theorem

Here are three more examples that show how the right angle at the point of contact is used.

✏️ Example 4 — Tangents at the ends of a diameter

Prove that the tangents drawn at the two ends of a diameter of a circle are parallel.

💡 Show solution▶

Given $AB$ is a diameter of a circle with centre $O$. The tangents at $A$ and $B$ are $l$ and $m$.

Right angles $OA$ and $OB$ lie along the diameter $AB$. By Theorem 10.1, $l\perp AB$ at $A$ and $m\perp AB$ at $B$.

Parallel The line $AB$ is a transversal that makes equal corresponding angles ($90^\circ$ each) with $l$ and $m$. Hence $l\parallel m$. $\blacksquare$

✏️ Example 5 — A road through a circular park

A circular park has radius $17$ m. A straight road passes at a perpendicular distance of $8$ m from the centre of the park. Is the road a secant, a tangent or outside the park? How long is the part of the road inside the park?

💡 Show solution▶

Compare $d=8\lt r=17$, so the road is a secant: it cuts the park boundary at two points.

Half the chord Let the foot of the perpendicular from the centre be $M$, and let one end of the road inside the park be $A$. In right-angled $\triangle OMA$: $MA^2=17^2-8^2=289-64=225$, so $MA=15$ m.

Length The perpendicular from the centre bisects the chord, so the road inside the park is $2\times15=30$ m long.

Check $8^2+15^2=64+225=289=17^2$ ✓.

✏️ Example 6 — A tangent length with an unknown

A circle has radius $10$ cm. From an outside point $T$, a tangent $TP$ is drawn, touching the circle at $P$. The distance $OT$ is $2$ cm more than the length $TP$. Find $TP$ and $OT$.

💡 Show solution▶

Unknown Let $TP=x$ cm. Then $OT=(x+2)$ cm.

Pythagoras $\angle OPT=90^\circ$, so $OT^2=OP^2+TP^2$: $(x+2)^2=10^2+x^2$.

Solve $x^2+4x+4=100+x^2$, so $4x=96$ and $x=24$.

Answer $TP=24$ cm and $OT=26$ cm.

Check $10^2+24^2=100+576=676=26^2$ ✓.

Exercises

In each part $O$ is the centre of the circle, $P$ the point of contact and $T$ a point on the tangent at $P$.

1 The radius is $9$ cm and $PT=12$ cm. Find $OT$.
2 $OT=25$ cm and the radius is $7$ cm. Find $PT$.
3 $\angle POT=65^\circ$. Find $\angle OTP$.
4 $OT=10$ cm and $PT=8$ cm. Find the radius.
5 $\angle OTP=32^\circ$. Find $\angle TOP$.
6 $PT=15$ cm and $OT=17$ cm. Find the radius, and the area of $\triangle OPT$.
7 $PT=OP$. Find $\angle OTP$.

💡 Show solutions▶

1 $\angle OPT=90^\circ$, so $OT^2=9^2+12^2=81+144=225$ and $OT=15$ cm.

2 $PT^2=25^2-7^2=625-49=576$, so $PT=24$ cm.

3 $\angle OTP=180^\circ-90^\circ-65^\circ=25^\circ$.

4 $OP^2=OT^2-PT^2=100-64=36$, so the radius is $6$ cm.

5 $\angle TOP=90^\circ-32^\circ=58^\circ$.

6 $OP^2=17^2-15^2=289-225=64$, so the radius is $8$ cm. Area $=\tfrac12\times8\times15=60$ cm$^2$.

7 $\triangle OPT$ is right-angled at $P$ with $OP=PT$, so it is isosceles and the two acute angles are equal: $\angle OTP=90^\circ\div2=45^\circ$.

Secant, tangent or neither? ($d$ is the perpendicular distance of the line from the centre, $r$ the radius.)

1 $r=13$ cm, $d=5$ cm. Classify the line and find the length of the chord it cuts off.
2 $r=6.5$ cm, $d=6.5$ cm. Classify the line.
3 $r=8$ cm, $d=9$ cm. Classify the line.
4 For the circle $x^2+y^2=25$ (centre the origin), classify the lines $y=3$, $y=5$ and $y=6$, and give the common points where they exist.
5 A circle has radius $2.4$ cm. For which distances $d\ge0$ is a line at distance $d$ from the centre a secant? For which is it outside the circle?
6 A line meets a circle of radius $7.5$ cm at exactly one point. How far is the line from the centre?
7 A chord of length $8$ cm is drawn in a circle of radius $5$ cm. How far is it from the centre?
8 How many tangents can be drawn to a circle through (a) a point on the circle, (b) a point inside the circle?

💡 Show solutions▶

1 $d\lt r$, so a secant. Half the chord $=\sqrt{13^2-5^2}=\sqrt{144}=12$ cm, so the chord is $24$ cm.

2 $d=r$: a tangent.

3 $d\gt r$: the line lies outside the circle (no common point).

4 The distance of $y=k$ from the origin is $|k|$ and $r=5$. $y=3$: secant, meeting the circle where $x^2+9=25$, i.e. at $(4,3)$ and $(-4,3)$. $y=5$: tangent, touching at $(0,5)$. $y=6$: $6\gt5$, no common point.

5 Secant when $0\le d\lt2.4$ cm; outside the circle when $d\gt2.4$ cm (tangent when $d=2.4$ cm).

6 Exactly one common point means a tangent, so $d=r=7.5$ cm.

7 The perpendicular from the centre bisects the chord, so half the chord is $4$ cm. Distance $=\sqrt{5^2-4^2}=\sqrt9=3$ cm.

8 (a) exactly one tangent (perpendicular to the radius at that point); (b) none, because every line through a point inside the circle cuts the circle at two points.

III

Concentric circles, proofs and distances.

1 Two concentric circles have radii $13$ cm and $5$ cm. Find the length of the chord of the larger circle that touches the smaller one.
2 Concentric circles of radii $10$ cm and $6$ cm: find the same chord length.
3 A chord of length $30$ cm of a circle touches a smaller concentric circle of radius $8$ cm. Find the radius of the larger circle.
4 Prove that the perpendicular to a tangent at its point of contact passes through the centre of the circle.
5 A point $T$ is $13$ cm from the centre $O$ of a circle of radius $5$ cm. Find the length of the tangent from $T$ and the shortest distance from $T$ to the circle.
6 Prove that a line through a point $P$ of a circle, perpendicular to the radius $OP$, is a tangent to the circle.
7 A chord $AB$ of a circle with centre $O$ touches a smaller concentric circle at $P$. Prove that $AP=PB$.

💡 Show solutions▶

1 Half the chord $=\sqrt{13^2-5^2}=\sqrt{144}=12$ cm; chord $=24$ cm.

2 Half the chord $=\sqrt{10^2-6^2}=\sqrt{64}=8$ cm; chord $=16$ cm.

3 The point of contact bisects the chord, so half the chord is $15$ cm. Radius $=\sqrt{15^2+8^2}=\sqrt{289}=17$ cm.

4 Let the tangent touch the circle at $P$. By Theorem 10.1, $OP$ is perpendicular to the tangent at $P$. Through the point $P$ there is only one line perpendicular to the tangent, so that line is $OP$ itself, and it passes through $O$. $\blacksquare$

5 Tangent length $=\sqrt{13^2-5^2}=12$ cm. The nearest point of the circle to $T$ is where $TO$ meets the circle, at distance $13-5=8$ cm from $T$.

6 Let $Q$ be any point of the line other than $P$. In right-angled $\triangle OPQ$ (right angle at $P$), $OQ$ is the hypotenuse, so $OQ\gt OP=r$. Hence $Q$ is outside the circle, and $P$ is the only common point: the line is a tangent. $\blacksquare$

7 $AB$ is a tangent to the smaller circle at $P$, so $OP\perp AB$ (Theorem 10.1). In the right-angled triangles $OPA$ and $OPB$: $OA=OB$ (radii of the larger circle), $OP$ is common, and $\angle OPA=\angle OPB=90^\circ$. They are congruent by RHS, so $AP=PB$. $\blacksquare$

Applications and a challenge.

1 A circular park has radius $25$ m. A straight road passes $7$ m from its centre. How long is the part of the road inside the park?
2 A satellite is $500$ km above the Earth's surface. Taking the Earth as a sphere of radius $6400$ km, how far is the horizon from the satellite — that is, how long is the straight line of sight from the satellite that just touches the Earth? Give the answer to $1$ decimal place.
3 A circular pond has radius $10$ m. A straight path touches the pond at $P$. A lamp post $T$ stands on the path $24$ m from $P$. Find the distance from the lamp post to the centre of the pond, and to the nearest point of the water's edge.
4 A stone is whirled on a string $1.5$ m long and released. It then travels $2$ m in a straight line. How far is it from the centre of the circle at that moment?
5 Challenge. A circle has centre $O$ and radius $3$ cm. The tangent from $T$ has length $4$ cm. The line $TO$ meets the circle at $A$ (nearer to $T$) and $B$. Find $TA$ and $TB$.

💡 Show solutions▶

1 Half the road $=\sqrt{25^2-7^2}=\sqrt{576}=24$ m. Length inside $=48$ m.

2 The line of sight is a tangent, so it is perpendicular to the Earth's radius at the horizon point. The centre-to-satellite distance is $6400+500=6900$ km. Horizon distance $=\sqrt{6900^2-6400^2}=\sqrt{47\,610\,000-40\,960\,000}=\sqrt{6\,650\,000}\approx2578.8$ km.

3 $OT=\sqrt{10^2+24^2}=\sqrt{676}=26$ m. The nearest point of the water's edge is on the line $OT$, at $26-10=16$ m from the lamp post.

4 The stone leaves along the tangent, so the distance from the centre is $\sqrt{1.5^2+2^2}=\sqrt{2.25+4}=\sqrt{6.25}=2.5$ m.

5 $OT=\sqrt{3^2+4^2}=5$ cm. $TA=OT-OA=5-3=2$ cm and $TB=OT+OB=5+3=8$ cm.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marks

How this is examined: "tangent $\perp$ radius" is a standard 3-mark proof. Short questions give a tangent figure and ask for a length (Pythagoras at the point of contact), an angle (angle sum of the right-angled triangle), or the chord of a larger concentric circle that touches a smaller one. One-mark items test the definitions: how many common points a tangent has, or the angle between tangent and radius.

E1. Board · 1 mark $PQ$ is a tangent at $P$ to a circle with centre $O$. If $OP=3$ cm and $OQ=5$ cm, then $PQ=$ (A) $3$ cm (B) $4$ cm (C) $\sqrt{34}$ cm (D) $8$ cm

💡 Show solution▶

Idea $\angle OPQ=90^\circ$, so $PQ^2=OQ^2-OP^2=25-9=16$ and $PQ=4$ cm.

Trap (C) $\sqrt{34}$ comes from adding $3^2+5^2$, as if $OQ$ were a leg; but $OQ$ is the hypotenuse. (D) $8=3+5$ adds the lengths without squaring.

Answer (B).

E2. Board · 1 mark Assertion (A): A line that meets a circle at two points can be a tangent to the circle. Reason (R): A tangent to a circle meets it at exactly one point.
(A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check R By definition (Definition 10.2) a tangent has exactly one common point with the circle. True.

Check A A line with two common points is a secant. So it cannot be a tangent: A is false.

Answer (D).

E3. Board · 2 marks $PT$ is a tangent at $P$ to a circle with centre $O$ and radius $4$ cm. If $\angle OTP=30^\circ$, find $OT$ and $PT$.

💡 Show solution▶

OT $\angle OPT=90^\circ$, so $\sin30^\circ=\dfrac{OP}{OT}$. Thus $\dfrac12=\dfrac4{OT}$ and $OT=8$ cm. (1 mark)

PT $PT^2=OT^2-OP^2=64-16=48$, so $PT=4\sqrt3\approx6.93$ cm (with $\sqrt3\approx1.732$). (1 mark)

Check $\tan30^\circ=\dfrac{4}{4\sqrt3}=\dfrac1{\sqrt3}$ ✓.

E4. Board · 3 marks Two concentric circles have radii $17$ cm and $15$ cm. Find the length of the chord of the larger circle which is a tangent to the smaller circle.

💡 Show solution▶

Figure Let the chord $AB$ touch the smaller circle at $P$; then $OP\perp AB$, $OP=15$ cm, $OA=17$ cm. (1 mark)

Half the chord $AP^2=OA^2-OP^2=289-225=64$, so $AP=8$ cm. (1 mark)

Chord The perpendicular from the centre bisects the chord, so $AB=2\times8=16$ cm. (1 mark)

E5. Board · 3 marks Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

💡 Show solution▶

Given, to prove (1 mark) A circle with centre $O$ and a tangent $XY$ touching it at $P$. To prove: $OP\perp XY$.

Proof (2 marks) Take any point $Q$ on $XY$ other than $P$ and join $OQ$. Since $XY$ meets the circle only at $P$, the point $Q$ lies outside the circle, so $OQ\gt OP$. This is true for every $Q\ne P$ on $XY$, so $OP$ is the shortest of all segments from $O$ to the line $XY$. The shortest segment from a point to a line is the perpendicular. Hence $OP\perp XY$. $\blacksquare$

🔬 Research–Teaching NexusMini project · optional, about 30 minutes

Research question. If you place a ruler at different distances from the centre of a circle you have drawn, how does the number of common points — and the length of the chord — depend on the distance? Does the ruler become a tangent exactly when its distance from the centre equals the radius?

Why it matters. Engineers who design belt drives, gear and pulley systems, road bends and railway curves rely on the fact that a straight part of a path joins a circular part smoothly only along a tangent, at a right angle to the radius. A cutter that is slightly too close to a round shaft removes a chord of material; the maths of chord length decides how much.

What to do.

With a compass, draw a circle of radius $4$ cm on plain paper and mark its centre $O$.
Lay a ruler so that its edge is $3$ cm from $O$ (measure with a set-square from $O$ to the edge, perpendicular to the edge). Mark the two points where the edge crosses the circle and measure the chord between them.
Repeat for distances $d=2$ cm, $3.5$ cm, $4$ cm and $5$ cm. For each, record the number of common points and the chord length, and compare with the formula $2\sqrt{r^2-d^2}$.
At $d=4$ cm, mark the point of contact and measure the angle between the radius and the ruler's edge with a protractor.
Write two sentences explaining your table in terms of Theorem 10.1.

Tools: Compass, ruler, set-square, protractor, plain paper, pencil

Hand in: Your drawing, a table of $d$, number of common points, measured chord, calculated chord and the percentage difference, and your explanation.

Going further. Predict the distance $d$ at which the chord is exactly $6$ cm for $r=4$ cm, and test your prediction by drawing. (Expected: $d=\sqrt{16-9}=\sqrt7\approx2.65$ cm.)

Teacher/parent note: at $r=4$ cm the expected chords are $2\sqrt{16-4}\approx6.93$ cm for $d=2$ cm, $2\sqrt7\approx5.29$ cm for $d=3$ cm, $2\sqrt{3.75}\approx3.87$ cm for $d=3.5$ cm, $0$ at $d=4$ cm. Measured values within about 1–2 mm are good. The common error is measuring $d$ along a slanted line instead of the perpendicular.

🧪 Self-check quiz

0/5 correct

Q1. $PT$ is a tangent at $P$ to a circle with centre $O$. If $OP=5$ cm and $PT=12$ cm, then $OT=$

Q2. A circle has radius $6$ cm. A line is $6$ cm from its centre (perpendicular distance). The line is

Q3. $PT$ is a tangent at $P$ to a circle with centre $O$, and $\angle TOP=55^\circ$. Then $\angle OTP=$

Q4. Two concentric circles have radii $10$ cm and $6$ cm. A chord of the larger circle touches the smaller circle. Its length is

Q5. Which statement is false?

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