In the figure; P and Q are the points of intersection of two circles with centres O & O'. If straight line APB and CQD are parallel to OO', then what is the ratio of OO' and AB?

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Solution

The correct option is B
1:2

Draw $O M ⊥ A P$ & $O^{'} N ⊥ P B$ .

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

OO' = MN

We know perpendicular from the Centre of the circle to the chord bisects the chord, therefore:

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

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

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

$∴ O O^{'} : A B = 1 : 2$

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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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is drawn parallel to

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through

A

(o r B)

intersecting the circles at

P

and

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