The lengths of tangents from a fixed point to three circles of coaxial system are ${t_1},{t_2},{t_3}$ and if $P, Q$ and $R$ be the centres, then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that

The two circles ${x^2} + {y^2} - 4y = 0$ and ${x^2} + {y^2} - 8y = 0$

If a circle $C,$  whose radius is $3,$ touches externally the circle, $x^2 + y^2 + 2x - 4y - 4 = 0$ at the point $(2, 2),$  then the length of the intercept cut by circle $c,$  on the $x-$ axis is equal to

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

The equation of a circle passing through origin and co-axial to circles ${x^2} + {y^2} = {a^2}$ and ${x^2} + {y^2} + 2ax = 2{a^2},$ is