From a point $P$ which is at a distance of $13 \, cm$ from the centre $O$ of a circle of radius $5 \, cm$,the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is (in $cm^2$)

In quadrilateral $ABCD$,$m \angle D = 90^\circ$. $A$ circle with center $O$ and radius $r$ touches its sides $AB, BC, CD,$ and $DA$ at points $P, Q, R,$ and $S$ respectively. If $BC = 38, CD = 25,$ and $BP = 25,$ find the radius $r$ of the circle.

$AB$ is a diameter and $AC$ is a chord of a circle with centre $O$ such that $\angle BAC = 30^{\circ}$. The tangent at $C$ intersects the extended $AB$ at a point $D$. Prove that $BC = BD$.

At one end $A$ of a diameter $AB$ of a circle of radius $5 \, cm$,a tangent $XY$ is drawn to the circle. The length of the chord $CD$ parallel to $XY$ and at a distance of $8 \, cm$ from $A$ is (in $cm$):

$\odot(O, 41)$ and $\odot(O, 9)$ are concentric circles. The chord $\overline{AB}$ of $\odot(O, 41)$ touches $\odot(O, 9)$ at point $M$. Then $AB = \ldots$

In the figure,if $PQR$ is the tangent to a circle at $Q$ whose center is $O$,$AB$ is a chord parallel to $PR$,and $\angle BQR = 70^{\circ}$,then $\angle AQB$ is equal to: (in $^{\circ}$)