The following is a problem from the chapter "Circles":
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 values of $p$
(B) all except one value of $p$
(C) all except two values of $p$
(D) exactly one value of $p$
My Attempt:
The family of circles passing though the point of intersection of the two given circles is given by the following equation:
$$(x^2+y^2+3x+7y+2p-5)+\lambda(x^2+y^2+2x+2y-p^2)=0$$
One of the circles represented by the above equation passes through the point $(1,1)$. So,
$$(1^2+1^2+3(1)+7(1)+2p-5)+\lambda(1^2+1^2+2(1)+2(1)-p^2)=0$$
$$(7+2p)+\lambda(6-p^2)=0$$
We know that there exists only one circle which passes through three fixed points. So, there must exist a unique value of $\lambda$ for the circle to exist. The final equation can be converted in such a way we obtain an expression for $\lambda$ as follows:
$$\lambda=\frac{2p+7}{p^2-6}$$
So for $\lambda$ to exist $p^2-6\neq0$ or $p\neq \pm \sqrt6$
So from this, I concluded that there is a circle passing through the points P,Q, and $(1,1)$ for all except two values of $p$ i.e., option (C). But surprisingly the answer given in my book is (A) all values of $p$. I checked the solution, but there is no reason specified for this.
The following image is the solution from one of the web resources for the same problem (which is of no use):
Almost all resources state the answer for the given problem is "all values of $p$". So clearly, I think I've made a mistake. But I am unable to figure where I went wrong. Kindly explain which option is correct and the reason for it.
Thank you in advance.
Your mistake is here:
This omits the circle $x^2+y^2+2x+2y-p^2=0$ (corresponding to $\lambda=\infty$ in your equation). And in fact if you put $p=\pm\sqrt 6$, it is easy to check that this circle passes through $P,Q$ and $(1,1)$ as long as $P$ and $Q$ exist (i.e. as long as the original two circles intersect).