The quadratic equation whose roots are the coordinates of the circumcentre of the triangle formed by the points $(-2,-1),(6,-1)(2,5)$ is

Equation of perpendicular bisector of line joining points $A(-2,-1)$ and $B(6,-1)$ is $[\because$ mid-point of $A B$ is $(2,-1)]$
and similarly equation of perpendicular bisector of line joining points $B(6,-1)$ and $C(2,5)$ is $[\because$ mid-point of $B C$ is $(4,2)]$ $$ y-2=\frac{4}{6}(x-4) $$
Now, point of intersection of perpendicular bisector is the circumcentre of $\triangle A B C$, so On solving Eqs. (i) and (ii), we get $C\left(2, \frac{2}{3}\right)$. It is given that quadratic equation has roots are 2 and $\frac{2}{3}$, so equation of required quadratic equation is $x^2-\left(2+\frac{2}{3}\right) x+2\left(\frac{2}{3}\right)=0$ $\Rightarrow 3 x^2-8 x+4=0$ Hence, option (c) is correct.