Question

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 $p$

Answer 1

Answer: (A)

all except one value of $p$

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

$S_1:x^2+y^2+3x+7y+2p-5=0$ and $S_2:x^2+y^2+2x+2y-p^2=0$ is

$S_1-S_2=0\Rightarrow x+5y+2p-5+p^2=0....\left(i\right)$

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.

$1+5+2p-5+p^2\ne 0$

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

$\therefore p\ne -1$

Thus for all values of $p$ except '$-1$' there is a circle passing through P, Q and $(1,1).$