8 min readConcept sheet3-mark proof, or the proof inside a 5-mark question; 1 and 2-mark uses

Theorem: the tangent is perpendicular to the radius

The tangent at any point of a circle is perpendicular to the radius through the point of contact. Proof: OP is the shortest distance from O to the tangent.

StatementThe tangent at any point of a circle ⟂ the radius through the point of contact

Key idea of the proofEvery other point of the tangent lies outside the circle, so OP is the shortest distance

Use in problemsA right angle at the point of contact, so use Pythagoras theorem

Tangent length√(d² − r²), where d = distance of the point from the centre

The theorem

Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact. Given: a circle with centre O and a tangent XY at the point P. To prove: OP ⟂ XY. This theorem gives a right angle at every point of contact. So you can use Pythagoras theorem in tangent problems. The line through P perpendicular to the tangent passes through the centre. It is called the normal to the circle at P.

How to draw the figure for the proof

Draw a circle and mark its centre O.
Draw a line XY that touches the circle at P.
Join OP.
Mark any other point Q on XY and join OQ.
Mark R, the point where OQ crosses the circle.

Proof

1
Take any point Q on XY other than P, and join OQ.
2★
Q must lie outside the circle. If Q were inside or on it, XY would meet the circle twice and be a secant, not a tangent.
3
Let OQ meet the circle at R. Then OQ = OR + RQ, so OQ > OR.
4
OR = OP (radii of the same circle). So OQ > OP.
5
This is true for every point Q of XY except P. So OP is the shortest distance from O to XY.
6
The shortest segment from a point to a line is perpendicular to it. So OP ⟂ XY. Hence proved.

★ key step

Finding the length of a tangent

Problem

The radius of a circle with centre O is 7 cm. A point Q is 25 cm from O. QP is a tangent from Q, touching the circle at P. Find QP.

Working

Theorem: tangent ⟂ radius, so ∠OPQ = 90°.
In right ΔOPQ, by Pythagoras theorem, OQ² = OP² + QP².
25² = 7² + QP², so QP² = 625 − 49 = 576.
QP = √576 = 24 cm.

Answer

QP = 24 cm

Word problem: a path touching a garden

Problem

A circular garden has radius 15 m. A straight path touches the garden at P. A lamp post L on the path is 20 m from P. How far is the lamp post from the centre O of the garden?

Working

The path is a tangent at P, so OP ⟂ PL and ∠OPL = 90°.
By Pythagoras theorem, OL² = OP² + PL².
OL² = 15² + 20² = 225 + 400 = 625.
OL = √625 = 25 m.

Answer

The lamp post is 25 m from the centre.

In real life

When a bullock cart wheel rolls on a flat road, the spoke pointing straight down is perpendicular to the road at the point of contact. That is this theorem in action.

Exam angle

1 marks: “The tangent at P to a circle with centre O meets a line through O at Q. What is ∠OPQ?” 90°, because the tangent is perpendicular to the radius OP.

2 marks: “From a point P, the length of the tangent to a circle is 15 cm. P is 17 cm from the centre. Find the radius.” r² = 17² − 15² = 289 − 225 = 64, so r = 8 cm.

3 marks: “Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.” Figure. Take Q ≠ P on the tangent; Q lies outside, so OQ > OR = OP. OP is the shortest distance from O to the line, so OP ⟂ the tangent.

Remember it

Radius meets tangent at 90°. Picture a wheel's spoke standing straight up from the road.