Question

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer 1

Let us consider the lines be$l_1$ and $l_2$_.

Will consider that $O$ touches $l_1$ and $l_2$ at $M$ and $N$,

So,

$OM=ON$ (Radius of the circle)

Therefore,

The centre ”$O$” of the circle, has an equal distance from $l_1$ & $l_2$.

From $\Delta OPM$ & $OPN$,

$OM=ON$ (Radius of the circle)

$\angle OMP=\angle ONP$ (As, Radius is perpendicular to its tangent)

$OP=OP$ (Common sides)

Therefore,

$\Delta OPM=\Delta OPN$ (SSS congruence rule)

By C.P.C.T,

$\angle MPO=\angle NPO$

So, $l$ bisects $\angle MPN$.

Therefore, $O$ lies on the bisector of the angle between $l_1$ & $l_2$ .

Hence, we prove that the center of a circle touching two intersecting lines lies on the angle bisector of the lines.