In the figure, $P$ and $Q$ are the points of intersection of two circles with centres $O$ and $O^{'} .$ If straight lines, $A P B$ and $C Q D$ are parallel to $O O^{'},$ then what is the ratio of $O O^{'}$ and $A B ?$

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Solution

Draw $O M ⊥ A P$ and $O^{'} N ⊥ P B$ .

Now, quadrilateral $O O^{'} N M$ is a rectangle, since, all of its angles are $90^{\circ}$

$\Rightarrow O O^{'} = M N$

We know that, perpendicular from the centre of the circle to the chord bisects the chord, therefore

$A M = M P = \frac{1}{2} A P$ and

$P N = N B = \frac{1}{2} B P$

$\Rightarrow O O^{'} = M N = P M + P N$
$= \frac{A P}{2} + \frac{B P}{2}$
$= \frac{(A P + B P)}{2}$
$= \frac{A B}{2}$

$∴ O O^{'} : A B = 1 : 2$

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