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 $C_1$ 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 ............. $\mathrm{sq. \, units}$

Let the centre of a circle, passing through the point $(0,0),(1,0)$ and touching the circle $x^2+y^2=9$, be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k), 4\left(h^2+k^2\right)$ is equal to .............

If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then

The equation of the circle which passes through the origin, has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally, is

A circle $\mathrm{C}$ touches the line $\mathrm{x}=2 \mathrm{y}$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2 y-5=0$ at two points $\mathrm{P}$ and $\mathrm{Q}$ such that $\mathrm{PQ}$ is a diameter of $\mathrm{C}_{1}$. Then the diameter of $\mathrm{C}$ is :