If P and Q are points of intersection of circles {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0 and {x^2} + {y

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10-1.Circle and System of Circles

hard

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 except one value of $p$

B

all except two values of $p$

C

exactly one value of $p$

D

all values of $p $

(AIEEE-2009)

Solution

The radical axis, which in the case of the intersection of the circles is the common chord, of the

circles

$S_{1}: x^{2}+y^{2}+3 x+7 y+2 p-5=0$ and $S_{2}: x^{2}+y^{2}+2 x+2 y-p^{2}=0$ is

$S_{1}-S_{2}=0 \Rightarrow x+5 y+2 p-5+p^{2}=0 \ldots(i)$

If there is a circle passing through $P, Q$ and (1,1) it's necessary and sufficient that (1,1) doesn't lie on PQ, i.e.

$1+5+2 p-5+p^{2} \neq 0$

$\Rightarrow p^{2}+2 p+1 \neq 0 \Rightarrow(p+1)^{2} \neq 0$

$\therefore p \neq-1$

Thus for all values of $p$ except $^{\prime}-1^{\prime}$ there is a circle passing through $P, Q$ and (1,1)

Standard 11

Mathematics

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