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If P and Q are the points of intersectio...

If P and Q are the points of intersection of the circles `x^(2)+y^(2)+3x+7y+2p-5=0` and `x^(2)+y^(2)+2x+2y-p^(2)=0`, then there is a circle passing through P,Q, and (1,1) for

A

all values of p

B

all except one value of p

C

all except two values of p

D

exactly one value of p

Text Solution

Verified by Experts

The correct Answer is:
A

The equation of the circle passing through P and Q is
`(x^(2)+y^(2)+3x+7y+2p-5)+lambda(x^(2)+y^(2)+2x+2y-p^(2))=0...(i)` [Using : `S_(1)+lambdaS_(2)=0`]
It passes through (1,1) .
`:. (2p+7)+lambda(6-p^(2))=0rArr lambda=(2p+7)/(p^(2)-6)`
If `p=-(7)/(2)`, then `lambda=0` and equation (i) represents first circle.
If `p^(2)=6`, then equation (i) represents second circle. For all other values of p, equation represents some other circle. Hence, equation(i) represents a circle for all values of p.
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OBJECTIVE RD SHARMA ENGLISH-CIRCLES-Section I - Solved Mcqs
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