$\overline{PA}$ is a tangent to $\odot(O, r)$ drawn from a point $P$ outside a circle. If $m\angle AOP = 40^\circ$,then $m\angle OPA = \ldots$ (in $^\circ$)

In $\Delta ABC$,$m \angle B = 90^{\circ}$. $A$ circle is inscribed in $\Delta ABC$ touching all its sides. If $AB = 16$ and $BC = 30$,find the radius of the circle.

In $Fig.$,$XY$ and $X'Y'$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersecting $XY$ at $A$ and $X'Y'$ at $B$. Prove that $\angle AOB = 90^{\circ}$.

$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $OP$,if $r = 12$ and $XP = 5$.

$A$ line passing through the centre of the circle $O$ intersects the tangent to the circle at $Q$. $P$ is the point of contact. If the radius of the circle is $9$ and $PQ = 40$,then find $OQ$.

$A$ tangent to a circle intersects the circle in $\ldots \ldots \ldots .$