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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:


a.

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.


b.

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.


c.

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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