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?

1 : 3

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1 : 2

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1 : 4

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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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Similar questions

Q. In the following figure: P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that
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Question 18
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