# 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 Option 1) all except one value of  p Option 2) all except two values of  p Option 3) exactly one value of  p Option 4) all values of p

As we learnt in

Family of circle -

$S+KL= 0$

- wherein

Equation of the family of circles passing through point of intersection $S=0 \, and\, line\, L=0$.

$x^{2}+y^{2}+3x+7y+2p-5 = 0$

${x^{2}+y^{2}+2x+2y-p^{2}=0}\\ (-)$

$\overline {{x+5y+\left ( p^{2}+2p-5 \right ) = 0}}$

Equation of common chord family of circles is

$\left ( x^{2}+y^{2}+3x+7y+2p-5 \right )+ \lambda \left ( x+5y+p^{2}+2p-5 \right )= 0$

It passes through (1,1)

$\left ( 7+2p \right )+ \lambda \left ( 2p+p^{2}+1 \right )= 0$

$\lambda = \frac{-\left ( 2p+7 \right )}{\left ( p+1 \right )^{2}}$

So, $p\neq 1\:\:\: only\:case.$

Option 1)

all except one value of  p

This option is correct.

Option 2)

all except two values of  p

This option is incorrect.

Option 3)

exactly one value of  p

This option is incorrect.

Option 4)

all values of p

This option is incorrect.

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