The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be 

Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $\mathrm{I}$ with center at the point $A=(4,1)$, where $1<\mathrm{r}<3$. Two distinct common tangents $P Q$ and $S T$ of $C_1$ and $C_2$ are drawn. The tangent $P Q$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $S T$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $P Q$ and $S T$ are joined to form a line which meets the $x$-axis at a point $B$. If $A B=\sqrt{5}$, then the value of $r^2$ is

The circle passing through the intersection of the circles, $x^{2}+y^{2}-6 x=0$ and $x^{2}+y^{2}-4 y=0$ having its centre on the line, $2 x-3 y+12=0$, also passes through the point

The equation of the circle through the point of intersection of the circles ${x^2} + {y^2} – 8x – 2y + 7 = 0$, ${x^2} + {y^2} – 4x + 10y + 8 = 0$ and $(3, -3)$ is