For how many values of p , the circle x ^ 2 + y ^ 2 + 2x + 4y - p =...

We shall consider 3 cases.

Case I: When p = 0

(i.e., circle passes through origin)

Now, equation of circle becomes

${x}^{2} + {y}^{2} + 2x + 4y = 0$

Case II: When circle intersect x-axis at 2 distinct points and touches y-axis.

${g}^{2} - c > 0$ and ${f}^{2} = c$

$1 + p > 0$ and $p = - 4$ so, this case is not possible

Case III: When circle intersect y-axis at 2 distinct points and touches x-axis

Now, ${g}^{2} - c = 0$  and ${f}^{2} - c > 0$

$⇒ 1 - \left(- p\right) = 0$ and $4 - \left(- p\right) > 0$

$⇒ p = - 1$  and, $p > - 4$

$∴ p = - 1$ is possible.

$∴$ Finally, we conclude that $p = 0, - 1$.

$⇒$Two possible values of p.