Question

A circle $(x-\alpha {)}^2+(y-\beta {)}^2=8$ is inscribed in a parabola $y^2=4x$, then $[\alpha +\beta ]$ is ([.] represents greatest integer function)

Answer 1

By symmetry, circle must be centred on
x-axis, so, $\beta =0$

Now, normal to $y^2=4x$ at $p(m^2,-2m)$ is
$y=mx-2m-m^3$

$(\alpha ,0)$ must lie on it $\Rightarrow m(\alpha -2)-m^3=0$

$m^2=\alpha -2$

Now, $P(m^2,-2m)$ lies on the given circle,

$\Rightarrow (m^2-\alpha {)}^2+4m^2=8$

$\Rightarrow (\alpha -2-\alpha {)}^2+4(\alpha -2)=8\Rightarrow \alpha =3$

$\Rightarrow \alpha +\beta =3+0$