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:

Any circle passing through the intersection of $S_1=0$ and $S_2=0$ is $S_1+\lambda S_2=0$.The point $(1,1)$ lies on it:$[1+1+3+7+2p-5]+\lambda [1+1+2+2-p^2]=0$$[7+2p]+\lambda [6-p^2]=0$.For a circle to exist, we must find a finite $\lambda$. This is possible for all $p$ except when $6-p^2=0$ AND $7+2p\ne 0$. If $6-p^2=0$, then $7+2p=0$ would mean $(1,1)$ lies on the common chord.According to the source, the condition holds for all values of $p$.