Mathematics - Circle Question with Solution | TestHub

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 $f^2-c>0$

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

$\Rightarrow p=-1$  and, $p>-4$

$\therefore p=-1$ is possible.

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

$\Rightarrow$Two possible values of p.