Question

Prove that the length of two tangents drawn from an external point to a circle are equal.

Answer 1

Hint: Assume a circle of radius r with centre O. From a point P outside the circle, draw two tangents to the circle PQ and PR. Join the point O to point P, Q, and R. Now, use the concept of congruent triangles to prove that the two tangents are equal. Using this, we can solve this question.

“Complete step-by-step answer:”
Let us consider a circle of radius r and centre at point O. Let us consider a point P which is outside the circle. From this point, let us draw two tangent PQ and PR to the circle. Finally, join the point O to the point Q, point R, point P.

Let us consider triangle POR and triangle POQ.
In right – angle $\Delta POR$ and $\Delta POQ$,
PO = PO (since PO is common in both the triangles)
$\angle OQP=\angle ORP={{90}^{\circ }}$ (since PQ and PR are tangents to the circle)
OQ = OR = r (since both OQ and OR are the radius of the circle)
So, from the RHS congruence rule for congruent triangles, $\Delta POQ\cong \Delta POR$.
From corresponding parts of congruent triangles, we can say,
$PQ=PR$
Hence, we can say that the two tangents from an external point to the circle are equal.

Note: There is a possibility that one may write that the triangles are congruent by ASS congruence rule. But we must know that there is no such rule as the ASS congruence rule. Hence, writing this in our answer will be considered as an incorrect answer.