Further Geometry

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Year 10 Interactive Maths - Second Edition

Tangent to a Circle

A tangent to a circle is a straight line that touches a circle at one point only.

For example, PT is a tangent to the following circle at the point P. We call P the point of contact.

The line PT is a tangent to the circle centred at O.

Note:

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

Example 20

The lines PA and PB are two tangents to a circle centred at O. The unknown angles in triangle POA are u degrees and x degrees. The unknown angles in triangle POB are y degrees and v degrees.

If PA and PB are two tangents to a circle centred at O, prove that:

a. PA = PB
b. u = v
c. x = y

Proof:

In triangle POA and triangle POB, angle PAO = angle PBO {Each = 90 degrees}, PO = PO (A common side}, OA = OB {Radii}. Therefore, triangle POA is congruent to triangle POB {RHS}. Therefore, PA = PB {Corresponding sides of congruent triangles}, u = v {Corresponding angles of congruent triangles} and x = y {Corresponding angles of congruent triangles}. As required.

In general:

If PA and PB are two tangents to a circle centred at O, then:

the tangents to the circle from the external point P are equal
OP bisects the angle between the two tangents
OP bisects the angle between the two radii to the points of contact

The lines PA and PB are two tangents to a circle centred at O.

Example 21

Find the value of the pronumeral(s) in each of the following diagrams:

Tangent TP to the circle centred at O forms a triangle OPT. Two angles in the triangle of size x degrees and 56 degrees are shown.

Tangent PA to the circle centred at O forms a triangle OPA. The length of PA, OP and OA is x cm, 5 cm and 13 cm respectively.

PR and PQ are two tangents to the circle centred at O. Angle OPQ is a degrees, angle POQ is 75 degrees, angle OPR is b degrees and angle POR is x degrees.

Solution:

(a) Angle OPT = 90 degrees {As OP is perpendicular to TP}. In triangle OPT, x + 56 + 90 = 180 {Angle sum of a triangle}. Solving for x, we find x = 34.

Angle OPA = 90 degrees {As OP is perpendicular to PA}. By Pythagoras' Theorem in Triangle OPA, x squared + 5 squared = 13 squared and after solving for x, x = 12.

(c) x = 75 {As PO bisects angle QOR}. In triangle OPQ, a + 75 + 90 = 180 {Angle sum of a triangle}. Solving for a, we find a = 15. b = a = 15 {As PO bisects angle RPQ and a = 15}

Key Terms

tangent to a circle, point of contact

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