In the figure given below, $O$ and $O\text{'}$ are centres of two circles touching each other externally at the point $P$. The common tangent at $P$ meets a direct common tangent $AB$ at $M$. Prove that: $\angle APB=90°$

 

In the adjoining figure, $O$ is the centre of the circle. If $\angle OAB=40°$, then $\angle ACB$ is equal to

In the adjoining figure, $O$ is the centre of the circle. If $\angle BAO=60°$, then $\angle ADC$ is equal to

In the adjoining figure, $O$ is the centre of a circle. If the length of chord $PQ$ is equal to the radius of the circle, then $\angle PRQ$ is

In the adjoining figure, if $O$ is the centre of the circle then the value of $x$ is

From a point which is at a distance of $13 \mathrm{cm}$ from the centre $O$ of a circle of radius $5 \mathrm{cm}$, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is

In the adjoining figure, $PQ$ and $PR$ are tangents from $P$ to a circle with centre $O$. If $\angle POR=55°$, then $\angle QPR$ is

In the adjoining figure, $PA$ and $PB$ are tangents from point $P$ to a circle with centre $O$. If the radius of the circle is $5 \mathrm{cm}$ and $PA\perp PB$, then the length $OP$ is equal to