Question

Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above, do the following:

In the fig., XP and XQ are tangents from T to the circle with centre O and R is any point on the circle. If AB is a tangent to the circle at R, prove that XA +AR = XB + BR

Answer 1

$\angle$OQP and $\angle$ORP are right angles, because these are angles between the radii and tangents,

Now in right triangles $△$OQP and $△$ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, $△$ OQP $\cong$ $△$ORP (RHS)

This gives PQ = PR

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

XP = XQ

XA + AP = XB + BQ …..(i)

Also BQ = BR and AR = AP

Substituting in (i)

XA + AR = XB + BR

Hence proved.