Let $x=1$

Hence, we get

$(\alpha^{2}-2\alpha+1)^{51}$

$=[(\alpha-1)^{2}]^{51}$

$=[\alpha-1]^{102}$

Now sum of the coefficients is $0$.

Hence

$\alpha-1=0$

$\alpha=1$

Therefore the point $(\alpha,2\alpha^{2})$ becomes $(1,2)$

$x^{2}+y^{2}-4=0$ is the equation of the given circle.

Substituting the acquired point, we get

$1+4-4$

$1>0$

Hence the point lies outside the circle.