Question

In the given figure, O is the centre of a circle and two tangents KR, KS are drawn on the circle from a point K lying outside the circle. Prove that KR = KS.

Answer 1

Given $KR$ and $KS$ are tangents to circle $O$ from point $K$

To prove that: $KR=KS$.

Construction: Join $OK$ as shown in the figure

Proof:

Normal is perpendicular to the tangent at a point on the circle.

Hence, $\angle OSK=\angle ORK={90}^o$

$\cos \angle SOK=\frac{OS}{OK}$

$\cos \angle ROK=\frac{OR}{OK}$

All points on the circle are equidistant from the center.

$i.e.OR=OS$

From above, $\cos \angle SOK=\cos \angle ROK$

$\angle SOK=\angle ROK$

$\sin \angle SOK=\sin \angle ROK$

$\frac{SK}{OK}=\frac{RK}{OK}$

$\therefore KR=KS$

Hence proved.