$\overline{PA}$ is a tangent to $\odot(O, 8)$ drawn from a point $P$ outside the circle. If $m\angle AOP = 45^\circ$,then $AP = \ldots$

The tangent at a point $C$ of a circle and a diameter $AB$ when extended intersect at $P$. If $\angle PCA = 110^{\circ},$ find $\angle CBA$ [see $Fig.$]. (in $^{\circ}$)

$A$ circle with centre $O$ touches sides $\overline{AB}$,$\overline{BC}$,$\overline{CD}$ and $\overline{DA}$ of quadrilateral $ABCD$ at points $P, Q, R$ and $S$ respectively. Prove that $m \angle AOB + m \angle COD = 180^{\circ}$ and $m \angle AOD + m \angle BOC = 180^{\circ}$.

$\overline{PA}$ and $\overline{PB}$ are the tangents to $\odot(O, r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 70^\circ$,then $m \angle POB = \dots$ (in $^\circ$)

If two circles touch each other externally,then $\ldots \ldots \ldots \ldots$ common tangents can be drawn to them.

In the given figure,if $\angle AOB = 125^{\circ}$,then $\angle COD$ is equal to: (in $^{\circ}$)