The equation of a circle passing through points of intersection of the circles ${x^2} + {y^2} + 13x – 3y = 0$ and $2{x^2} + 2{y^2} + 4x – 7y – 25 = 0$ and point $(1, 1)$ is

The equation of radical axis of the circles ${x^2} + {y^2} + x – y + 2 = 0$ and $3{x^2} + 3{y^2} – 4x – 12 = 0,$ is

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

If $P$ and $Q$ are the points of intersection of the circles ${x^2} + {y^2} + 3x + 7y + 2p – 5 = 0$ and ${x^2} + {y^2} + 2x + 2y – {p^2} = 0$ then there is a circle passing through $P, Q$ and $(1, 1)$ for:

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 – 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is …………. $\mathrm{sq. \, units}$