Tangent properties — right angles and equal lengths — Mathematics

Before this

Angles in a semicircle and cyclic quadrilaterals
Congruent triangles · coming soon
Pythagoras theorem · coming soon

This leads to

The alternate segment theorem
Tangents and normals to a circle · coming soon
Differentiation from first principles · coming soon

A wheel on a road

A bicycle wheel rolls along a flat road. At every moment, the wheel touches the road at one point. Draw the spoke from the centre of the wheel down to that point. The spoke stands at exactly 90° to the road.

It does not matter how big the wheel is. The angle is always 90°. Something in the geometry forces it.

Here is a second puzzle. Stand at a point outside a round pond. There are exactly two straight lines from you that just touch the edge of the pond, one on each side. Walk along each line to the point where it touches the edge. Which walk is longer?

Most students guess that it depends on where you stand. It does not. The two walks are always the same length. This lecture shows why both facts are forced, and how the first one makes the second true.

A line that touches a circle

A straight line and a circle can meet in three ways. The line can cross the circle at two points. It can miss the circle completely. Or it can touch the circle at exactly one point.

A line can cross a circle twice (a secant), touch it once (a tangent) or miss it. The tangent touches at T, and the radius OT meets it at a right angle.

A line that crosses the circle at two points is called a secant. The part of it inside the circle, between the two points, is a chord. So a chord is a piece of a secant.

A line that meets the circle at exactly one point is called a tangent. The point where it touches is the point of contact (some books say "point of tangency"). In the wheel, the road is a tangent.

What decides the case is the distance from the centre O to the line. If it is less than the radius r, the line cuts in and meets the circle twice. If it is more than r, the line misses. Only a distance of exactly r gives one point. The whole proof rests on this idea.

Why the radius meets the tangent at 90°

This is the hard part of the lecture. Spend time here.

Start with a tangent that touches the circle at T. Draw the radius OT. We want to show that the angle between OT and the tangent is 90°.

We ask what would go wrong if it were not 90°. This is a proof by contradiction: suppose the opposite, and show that it leads to something impossible.

So suppose a line passes through T, and it is not at 90° to OT. From O, drop a perpendicular to the line, and call its foot F. Because the angle at T is not 90°, F is not the same point as T.

Suppose a line through T is not at 90° to OT. The perpendicular from O meets the line at F, and OF is shorter than the radius. Going the same distance past F gives T′, a second point on the circle.

Look at the triangle OFT. It has a right angle at F. In a right-angled triangle, the side opposite the right angle is the longest side. That side is OT. So OF is shorter than OT. This means that OF is shorter than the radius.

Now use the idea from the last section. The distance from O to the line is OF, and it is less than r. So the line cuts into the circle. It must meet the circle at a second point.

To find the second point, go from T to F, then the same distance again past F, to T′. The triangles OFT and OFT′ have a right angle at F, the same side OF, and FT = FT′. So they are congruent, and OT′ = OT = r. This means T′ is on the circle too.

So the line meets the circle at T and T′. It is a secant, not a tangent, which contradicts what we started with. The only escape is for F to be T itself. So the radius OT is at 90° to the line.

That is it. That is the whole idea. A line that is not at 90° to the radius always dips inside the circle. Only the line at 90° stays outside everywhere except at T.

Tangent and radius

A tangent to a circle is perpendicular to the radius at the point of contact.

If the tangent touches the circle at T and O is the centre, the angle between OT and the tangent is 90°.

Euclid proves this in his Elements, Book III, Proposition 18, by the same idea: the perpendicular from the centre would be shorter than a radius. In the explorer below, try the argument yourself.

Explore — turn a line about the point T

The circle has radius 5 cm. Turn the line with the slider or the − / + buttons. Watch the dashed distance OF and the second point T′. Then press 90° and read what changes.

Jump to:

Angle between the radius OT and the line: 60°

This table shows a line through T at five angles to the radius OT, on a circle of radius 5 cm. OF is the distance from O to the line, and TT′ is the distance from T to the second meeting point.

Angle with OT	OF (cm)	TT′ (cm)	Meeting points
30°	2.50	8.66	2
60°	4.33	5.00	2
90°	5.00	0	1 (tangent)
120°	4.33	5.00	2
150°	2.50	8.66	2

Every angle except 90° gives a second point T′, because OF is less than r. As the angle gets closer to 90°, T′ slides towards T. At exactly 90°, the two points join into one.

Stop and think

A line passes through a point T on a circle, and it is at 90° to the radius OT. Must the line be a tangent? Or could it still cross the circle somewhere else?

Show the answer

It must be a tangent. Take any other point X on the line. The triangle OTX has a right angle at T. By Pythagoras, $O X^{2} = O T^{2} + T X^{2}$ . Since TX is not zero, $O X^{2}$ is bigger than $O T^{2}$ . So OX is longer than the radius, and X is outside the circle. Every point of the line except T is outside, so the line touches at T only. This backwards result is the converse. It is how you draw a tangent at a point: draw the radius, then draw a line at 90° to it.

Two tangents from one point

Now return to the pond. You stand at a point P outside a circle with centre O. Two tangents from P touch the circle at A and B. The lengths PA and PB are called the tangent lengths from P.

Draw the radii OA and OB, and join O to P. The figure splits into two triangles, OAP and OBP.

Two tangents from P touch the circle at A and B. The right angles at A and B, the equal radii and the shared side OP make the two triangles congruent, so PA = PB.

$∠ O A P = ∠ OBP = 9 0^{\circ}$ , because each tangent is perpendicular to the radius at its point of contact.
OP is a side of both triangles. It is the side opposite the right angle, the hypotenuse, in each of them.
OA = OB, because both are radii.

Two right-angled triangles with equal hypotenuses and one other pair of equal sides are congruent. This is the RHS test: right angle, hypotenuse, side. So triangle OAP is congruent to triangle OBP.

Congruent triangles match in every part. So the third sides are equal: PA = PB. The two walks round the pond are the same length.

Without the right angles, there would be no RHS and no proof. The first theorem makes the second one true.

Tangents from an external point

The two tangents drawn to a circle from a point P outside it have equal lengths: PA = PB.

The line OP bisects the angle APB and the angle AOB.

The second line also comes from the congruence. Matching angles are equal, so $∠ A PO = ∠ BPO$ and $∠ A OP = ∠ BOP$ . The figure is symmetrical about OP: fold the page along OP, and A lands on B.

You can also find the tangent length with numbers. In triangle OAP the right angle is at A, so Pythagoras gives

P A^{2} = O P^{2} - r^{2}

Nothing in this says which tangent you mean. That is another way to see that PA = PB.

Drawing the tangents from a point

To draw the tangents accurately, you need A and B: the points on the circle where $∠ O A P = 9 0^{\circ}$ . The last lecture showed that the points which see a segment at 90° lie on the circle with that segment as diameter. So draw the circle with diameter OP, centre M, the midpoint of OP. It crosses the first circle at A and B.

The dashed circle has diameter OP and centre M. It crosses the first circle at A and B, the two points of contact. O, A, P and B all lie on the dashed circle.

Join P to A and to B. The angle in a semicircle gives the right angles, and the converse above shows that the lines touch.

The figure shows something else. O, A, P and B all lie on the dashed circle. So OAPB is a cyclic quadrilateral. Its opposite angles at A and B are both 90°, and $9 0^{\circ} + 9 0^{\circ} = 18 0^{\circ}$ , exactly as the cyclic quadrilateral theorem says.

Stop and think

In the last lecture, a shape OAPB with a corner at the centre was not cyclic. Here, OAPB is cyclic. Is that a contradiction?

Show the answer

No. The two shapes are different. In the last lecture, P was on the original circle, and O was its centre, so O could not be on that circle. Here, P is outside, and the four points lie on a different circle, the one with diameter OP. "Cyclic" asks whether there is some circle through all four corners. Always check which circle you mean.

This gives the reason for a fact that many students learn as a rule. Since OAPB is cyclic, its other pair of opposite angles also adds up to 180°:

∠ A OB + ∠ A PB = 18 0^{\circ}

The angle sum of a quadrilateral gives the same result: 360° minus two right angles leaves 180°. Either reason is a full reason in an exam.

Worked examples

Worked Example 1 — Finding a distance

A circle has centre O and radius 5 cm. A tangent touches it at T. The point P is on the tangent, with PT = 12 cm. Find OP, and find how far P is from the circle.

Think first. A radius meets a tangent, so mark the right angle at once. It is at T, the point of contact.

Triangle OTP is right-angled at T. The hypotenuse is the side opposite T, which is OP. Pythagoras gives

O P^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169

So $OP = 169 = 13$ cm.

The nearest point of the circle to P is on OP, so P is $13 - 5 = 8$ cm from the circle.

Check. The hypotenuse must be the longest side, and 13 is longer than both 5 and 12. ✓

Worked Example 2 — Equal tangents and algebra

Two tangents from P touch a circle at A and B. PA = (3x + 1) cm and PB = (5x − 7) cm. Find x and the tangent length.

Think first. Two tangents from the same point are equal. That gives an equation.

3 x + 1 = 5 x - 7

Add 7 to both sides and subtract 3x from both sides. This gives $8 = 2 x$ , so $x = 4$ .

The tangent length is $3 \times 4 + 1 = 13$ cm.

Check. Use the other expression: $5 \times 4 - 7 = 13$ cm. ✓ The two lengths agree, as the theorem says they must.

Worked Example 3 — An angle at the centre

Tangents from P touch a circle with centre O at A and B. $∠ A PB = 6 4^{\circ}$ . Find $∠ A OB$ , $∠ A PO$ and $∠ A OP$ .

Think first. Mark the two right angles at A and B. Then the quadrilateral OAPB has three known angles.

The angles of a quadrilateral add up to 360°. So

∠ A OB = 36 0^{\circ} - 9 0^{\circ} - 9 0^{\circ} - 6 4^{\circ} = 11 6^{\circ}

OP bisects the angle at P, so $∠ A PO = 6 4^{\circ} \div 2 = 3 2^{\circ}$ . OP also bisects the angle at O, so $∠ A OP = 11 6^{\circ} \div 2 = 5 8^{\circ}$ .

Check. Triangle OAP has angles 90°, 32° and 58°, and $9 0^{\circ} + 3 2^{\circ} + 5 8^{\circ} = 18 0^{\circ}$ . ✓ Also $11 6^{\circ} + 6 4^{\circ} = 18 0^{\circ}$ , as the cyclic quadrilateral OAPB says.

Worked Example 4 — How far is the horizon?

Your eyes are 1.7 m above a calm sea. How far away is the horizon? Take the Earth as a sphere of radius 6371 km, and ignore the small bending of light by the air.

Think first. Your line of sight to the horizon just grazes the curved sea. So it is a tangent. The point where it touches is the horizon, H.

Let O be the centre of the Earth and E your eye. Then OH = 6 371 000 m, a radius. OE = 6 371 000 + 1.7 = 6 371 001.7 m. The tangent EH is at 90° to the radius OH, so

E H^{2} = O E^{2} - O H^{2}

Work it out on a calculator. The answer is $E H^{2} \approx 21661403$ m², so $E H \approx 4654$ m, which is about 4.7 km.

From a lighthouse 100 m high, the same method gives about 35.7 km.

Check. Multiply out the brackets: $O E^{2} - O H^{2} = 2 \times 6371000 \times 1.7 + 1. 7^{2}$ . The last term is tiny, and $2 \times 6371000 \times 1.7 \approx 4654$ . ✓

Three traps

Before the traps, find the mistake in a student's working.

Spot the error — a tangent length

One line of this working contains the mistake that the rest follows from. Tap the line you think is wrong. You can try again as often as you like.

This working finds a tangent length, and one line contains the mistake.

A tangent from P touches a circle with centre O at T. OT = 7 cm and OP = 25 cm. Find PT.
∠OTP = 90°, because the tangent is perpendicular to the radius at T.
So triangle OTP is right-angled at T.
The error: "The longest side is PT, so PT² = OT² + OP²." The right angle is at T, so the longest side is OP. The correct equation is PT² = OP² − OT².
So PT² = 625 − 49 = 576, and PT = 24 cm, which is shorter than OP, as it must be.

Trap 1 — The right angle in the wrong place

The right angle is at the point of contact, where the radius meets the tangent. It is never at the outside point P. So in triangle OTP, the hypotenuse is OP, the line from the centre to the outside point. A tangent length is always shorter than OP.

Trap 2 — Equal tangents from different points

PA = PB only when both tangents start at the same point P. Two tangents from different points can have any lengths. Before you write "equal tangents", find the point that both lines come from.

Trap 3 — A line that only looks like a tangent

A line in a diagram may look as if it touches the circle. Do not use 90° unless the question says it is a tangent, or you have proved it. "Looks like" is not a reason.

Try it yourself

Work each one on paper first. Draw a diagram for every question and mark the right angles. Open a hint only when you are stuck.

A tangent from P touches a circle with centre O at T. OP = 17 cm and the radius is 8 cm. Find the tangent length PT.

Hint 1
Where is the right angle? Which side is the hypotenuse?

Hint 2
The right angle is at T, so OP is the hypotenuse. Write $P T^{2} = 1 7^{2} - 8^{2}$ .

Full answer
$∠ OTP = 9 0^{\circ}$ (tangent perpendicular to radius). So $P T^{2} = O P^{2} - O T^{2} = 289 - 64 = 225$ , and $PT = 15$ cm. Check: $8^{2} + 1 5^{2} = 64 + 225 = 289 = 1 7^{2}$ . ✓ And 15 is shorter than the hypotenuse, 17. ✓
Tangents from P touch a circle with centre O at A and B. $∠ A OB = 14 0^{\circ}$ . Find $∠ A PB$ , $∠ OP A$ and $∠ P A B$ .

Hint 1
OAPB has two right angles. What is left for the angles at O and P?

Hint 2
Triangle PAB has PA = PB, so it is isosceles. Its two base angles are equal.

Full answer
$∠ A PB = 18 0^{\circ} - 14 0^{\circ} = 4 0^{\circ}$ (OAPB has two right angles, so the other two angles add up to 180°). OP bisects the angle at P, so $∠ OP A = 2 0^{\circ}$ . Triangle PAB is isosceles because PA = PB, so $∠ P A B = (18 0^{\circ} - 4 0^{\circ}) \div 2 = 7 0^{\circ}$ . Check: triangle OAB is isosceles with $∠ O A B = (18 0^{\circ} - 14 0^{\circ}) \div 2 = 2 0^{\circ}$ , and $∠ O A B + ∠ P A B = 2 0^{\circ} + 7 0^{\circ} = 9 0^{\circ} = ∠ O A P$ . ✓
Tangents from P touch a circle of radius 6 cm at A and B, and PA = 8 cm. Find OP. Then find the length of the chord AB.

Hint 1
OP comes from Pythagoras. For AB, use the symmetry: OP crosses AB at a point N, at 90°, and N is the midpoint of AB.

Hint 2
Find the area of triangle OAP in two ways: $\frac{1}{2} \times O A \times P A$ , and $\frac{1}{2} \times OP \times A N$ .

Full answer
$O P^{2} = 6^{2} + 8^{2} = 100$ , so OP = 10 cm. The figure is symmetrical about OP, so OP meets AB at its midpoint N, at 90°. The area of triangle OAP is $\frac{1}{2} \times 6 \times 8 = 24$ cm². It is also $\frac{1}{2} \times 10 \times A N$ . So $5 \times A N = 24$ , and AN = 4.8 cm. This means AB = 9.6 cm. Check: in triangle OAN, $O N^{2} = 6^{2} - 4. 8^{2} = 12.96$ , so ON = 3.6 cm. In triangle PAN, $P N^{2} = 8^{2} - 4. 8^{2} = 40.96$ , so PN = 6.4 cm. And $3.6 + 6.4 = 10 = OP$ . ✓
A triangle ABC has AB = 7 cm, BC = 8 cm and CA = 9 cm. A circle inside it touches all three sides. Find the distance from each corner to the two points of contact next to it.

Hint 1
Each corner is an outside point with two tangents to the circle. So each corner has one tangent length. Call them a, b and c for the corners A, B and C.

Hint 2
Side AB is made of the tangent from A and the tangent from B, so $a + b = 7$ . Write the other two equations and add all three.

Full answer
Each side is split by its point of contact into two tangent lengths, one from each end. So $a + b = 7$ , $b + c = 8$ and $a + c = 9$ . Adding gives $2 (a + b + c) = 24$ , so $a + b + c = 12$ . Then $c = 12 - 7 = 5$ , $a = 12 - 8 = 4$ and $b = 12 - 9 = 3$ . The distances are 4 cm from A, 3 cm from B and 5 cm from C. Check: $4 + 3 = 7$ , $3 + 5 = 8$ , $4 + 5 = 9$ . ✓ Notice that 12 is half the perimeter. Each tangent length is half the perimeter minus the side opposite that corner.

The bigger idea

Think again about the explorer. A line through T at an angle meets the circle again at T′. As the angle moves towards 90°, T′ slides towards T. The tangent is what the secant becomes when its two meeting points join into one.

That picture is much bigger than circles. A curve such as $y = x^{2}$ has no centre and no radius. But you can still draw a line through two of its points and let one slide towards the other. The line it settles on is the tangent to the curve, and its gradient tells you how steep the curve is at that one point. That is the starting idea of calculus. The circle is where you see it first, because here the answer is something you can prove with a triangle.

Key concepts in this lesson

A tangent meets a circle at exactly one point, the point of contact, while a secant crosses it at two points.
The tangent is perpendicular to the radius at the point of contact, because at any other angle the line cuts the circle twice.
Conversely, a line through a point on the circle at 90° to the radius there is a tangent, which is how a tangent is drawn.
The two tangents from an outside point P are equal in length, by RHS congruence, and OP is a line of symmetry that bisects the angles at P and at O.
OAPB is cyclic on the circle with diameter OP, so the angle at the centre and the angle between the tangents add up to 180°.