Question

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

Answer 1

Let O be the centre of a circle.

Let PA and PB are two tangents drawn from a point P, lying outside the circle . Join OA, OB, and OP.

We have to prove that PA = PB

In $\Delta$OAP and $\Delta$OPB,

OAP = OPB (Each equal to ${90}^{\circ }$)

(Since we know that a tangent at any point of a circle is perpendicular to the radius through the point of contact and hence, OA $\Lambda$ PA and OB $\Lambda$ PB)

OA = OB (Radii of the circle)

OP = PO (Common side)

Therefore, by RHS congruency criterion,

$\Delta$ OPA $\cong$ $\Delta$OPB

$\therefore$ By CPCT,

PA = PB

Thus, the lengths of the two tangents drawn from an external point to a circle are equal.