Two congruent circles of centres $\mathrm{O}$ and $\mathrm{O}\text{'}$ intersects each other at point $\mathrm{A}$ and $\mathrm{A}\text{'}$, then prove that $\angle \mathrm{AOB}=\angle \mathrm{AO}\text{'}\mathrm{B}$.

The length of two chords $AB$ and $CD$ of a circle of centre $O$ are equal and $\angle AOB=60°$ then, $\angle COD$ is

Suppose $S_1 \mathrm{a}\mathrm{n}\mathrm{d} S_2$ are two unequal circles; $AB$ and $CD$ are the direct common tangents to these circles. A transverse common tangent $PQ$ cuts $AB$ in $R$ and $CD$ in $S$. If $AB=10$ units, then $RS$ is -

On the circle with center $O$, points $A,B$ are such that $OA = AB$. A point $C$ is located on the tangent at $B$ to the circle such that $A$ and $C$ are on the opposite sides of the line $OB$ and $AB=BC$. The line segment $AC$ intersects the circle again at $F$. Then the ratio $\angle BOF:\angle BOC$ is equal to -

The two chords $AB$ and $CD$ of a circle are at equal distance from the centre $O$. If $\angle AOB=60°$ and $CD=6 cm$, then calculate the length of the radius of the circle.

In the given circle, with centre $O$, $\mathrm{K} \mathrm{and} \mathrm{L}$ are the mid-points of equal chords $\text{AB and CD}$ respectively. $\angle OLK=25°$, then the value of $\angle LKB$ is equal to