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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:
all except one value of $p$
all except two values of $p$
exactly one value of $p$
all values of $p $
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)
