Tangents: the right angle at the point of contact and equal tangent lengths

From a point 13 cm from the centre of a circle of radius 5 cm, both tangents are exactly 12 cm long.

A wheel standing on the road

Look at a bicycle standing on a flat road. The tyre touches the road at one point, so the road is a tangent line. The radius from the hub down to that point is vertical and the road is horizontal: they meet at a right angle. On a slope the radius to the contact point is no longer vertical, but it is still at right angles to the road. This is not a property of bicycles. For every circle and every tangent, the radius drawn to the point of tangency is perpendicular to the tangent.

Euclid proves it by contradiction in Book III, Proposition 18. Suppose the radius OT to the point of tangency were not perpendicular to the tangent. Drop a perpendicular from O to the tangent line instead, meeting it at a different point G. In triangle OGT the angle at G is a right angle, so the side opposite it, OT, is the longest side, and OG is shorter than OT. But OT is a radius. A point on the line closer to O than a radius lies inside the circle, and a tangent never enters the circle. So G cannot exist, and OT itself is the perpendicular. The converse holds too, as a corollary to Proposition 16: a line drawn at right angles to a radius at its outer end touches the circle.

Two tangents from one point are equal

From a point P outside a circle you can draw exactly two tangents, touching the circle at T and T′. The segments PT and PT′ have the same length. Join O to T, to T′ and to P. Triangles OTP and OT′P each have a right angle, at T and at T′. They share the side OP, and OT = OT′ because both are radii. By the Pythagorean theorem PT² = OP² − r² and PT′² = OP² − r², so PT = PT′. The same picture shows that OP cuts the angle TPT′ in half, since the two right triangles are mirror images. The length PT is called the length of the tangent from P.

The theorem of Pythagoras also gives the length. With OP = 13 cm and r = 5 cm, PT² = 169 − 25 = 144, so PT = 12 cm on both sides. Going the other way, if a tangent of 15 cm reaches a circle of radius 8 cm, the point is √(225 + 64) = √289 = 17 cm from the centre. In the quadrilateral PTOT′ the angles at T and T′ are 90° each, so the angle at P and the angle TOT′ add up to 180°. If the tangents meet at 50°, the angle between the two radii is 130°.

Drawing the tangents with a compass

To draw the tangents from P you need the points T where angle OTP is 90°. Unit 5 already found all such points: they lie on the circle with diameter OP. So find the midpoint M of OP, draw the circle with centre M through O and P, and it crosses the first circle at T and T′. Join P to each. Euclid’s own construction in Book III, Proposition 17 takes a different route, drawing a circle around O through P and a perpendicular from the first circle, but it ends with the same right angle at the point of contact.

Short answers from equal tangents

Equal tangents make some puzzles almost free. Draw the two tangents from P, each 10 cm long. Then draw a third tangent touching the circle somewhere on the near arc, cutting off a small triangle with P. Its perimeter is always 20 cm, wherever the third tangent touches. Each end of the third tangent is an outside point with two equal tangents, so the third side splits into two pieces that match the pieces of PT and PT′ it cuts off. The same idea works for a quadrilateral drawn around a circle: each corner sends two equal tangents, so the sums of opposite sides are equal. With sides 5, 6 and 9 in order, the fourth side is 5 + 9 − 6 = 8.