In a circle with center O- a chord PQ is such that OM- PQ meeting PQ at M- Then

The correct option is A OQ ^ 2 = OM ^ 2 + 1 4 PQ ^ 2 OM± PQ or OM bisects PQor MQ= 1 / 2 PQ or MQ^ 2 + 1 / 4 PQ ^ 2 By pythagoras theo ...

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Byju's Answer

Standard IX

Mathematics

Circle

In a circle w...

Question

In a circle with center $O$ , a chord $P Q$ is such that $O M \pm P Q$ meeting $P Q$ at $M$ . Then

{O Q}^{2} = {O M}^{2} + \frac{1}{2} {P Q}^{2}

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{O Q}^{2} = {O M}^{2} + \frac{1}{4} {P Q}^{2}

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{M Q}^{2} = {O M}^{2} - {O Q}^{2}

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{O M}^{2} = {M Q}^{2} - {O Q}^{2}

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Solution

The correct option is A ${O Q}^{2} = {O M}^{2} + \frac{1}{4} {P Q}^{2}$
$O M \pm P Q$ or OM bisects PQ
or $M Q = \frac{1}{2} P Q$
or $M Q^{2} + \frac{1}{4} {P Q}^{2}$
By pythagoras theorm,
${O Q}^{2} = {O M}^{2} + {M Q}^{2} = {O M}^{2} + \frac{1}{4} {P Q}^{2}$

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Similar questions

O

is any point inside a square

P Q R S

. Prove that

{O P}^{2} + {O R}^{2} = {O Q}^{2} + {O S}^{2}

If a circle and rectangular hyperbola $x y = c^{2}$ meets in 4 points $P, Q, R a n d S$ then $O P^{2} + O Q^{2} + O R^{2} + O S^{2} =$ ______ where r is the radius of the circle. O is the origin.

Q. In the figure,

A B

and

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are chords of a circle with center

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Point

M

lies on

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such that

O M

is perpendicular to

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and point

N

lies on

P Q

such that

O N

is perpendicular to

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Find the length of

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Q. If co-ordinates of

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and

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are

(a cos θ, b sin θ)

and

(- a sin θ, b cos θ)

respectively , then show that

O P^{2} + O Q^{2} = a^{2} + b^{2},

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Q. In a triangle OAB,

∠

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{OP}^{2} + {OQ}^{2} = \frac{5}{9} {AB}^{2}

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In a circle with center O, a chord PQ is such that OM±PQ meeting PQ at M. Then

The correct option is A OQ2=OM2+14PQ2OM±PQ or OM bisects PQor MQ=12PQor MQ2+14PQ2By pythagoras theorm,OQ2=OM2+MQ2=OM2+14PQ2

In a circle with center $O$ , a chord $P Q$ is such that $O M \pm P Q$ meeting $P Q$ at $M$ . Then

The correct option is A ${O Q}^{2} = {O M}^{2} + \frac{1}{4} {P Q}^{2}$
$O M \pm P Q$ or OM bisects PQ
or $M Q = \frac{1}{2} P Q$
or $M Q^{2} + \frac{1}{4} {P Q}^{2}$
By pythagoras theorm,
${O Q}^{2} = {O M}^{2} + {M Q}^{2} = {O M}^{2} + \frac{1}{4} {P Q}^{2}$