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Perpendicular from the Center to a Chord Bisects the Chord

In the figure...

Question

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
(i) OO’ = ½ AB
(ii) AB = CD

P and Q are points on intersection of two circles with centre O and O'. If straight line APB and CQD are parallel to OO' then AB = ___

P

is a point of intersection of two circles with centres

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and

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. If the straight line

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is parallel to

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Two circles whose centres are $O$ and $O^{'}$ intersect at $P$ . Through $P$ , a line parallel to $O O^{'}$ , intersecting the circle at C and D is drawn as shown. Then $C D = 2 O O^{'}$ .

Q. Two circles with centres

O

and

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intersect at two points

A

and

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A

line

P Q

is drawn parallel to

O O^{'}

through

A

(o r B)

intersecting the circles at

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and

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Prove that

P Q = 2 O O^{'}

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