Question

If $P$ and $Q$ are the points of intersection of the circle $x^2+y^2+3x+7y+2p=0$ and $x^2+y^2+2x+2y-p^2=0,$ then for what values of $p$ is there 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 value of $p$
• D. all exactly one value of $p$

Answer 1

Answer: (A)

All values of $p$

Equation of circle:

$\begin{array}{c}{x}^2+y^2+3x+7y+2p=\mathrm{0.............}\left(1\right)\\ x^2+y^2+2x+2y-p^2=\mathrm{0.............}\left(2\right)\end{array}$

For prove of intersection solving $\left(1\right)$ and $\left(2\right)$ we get,

$\begin{array}{c}{x}^2+y^2+3x+7y+2p=0\\ \underset{–}{x^2+y^2+2x+2y-p^2=0}\\ x+5y+2p+p^2=0\end{array}$

Putting $x=0$, $Y=\frac{-\left(2p+p^2\right)}{5}$

$Y=0$, $X=-\left(2p+p^2\right)$

Here equation is possible for any value of $p$.

Hence, this is the answer.