Let the circles $C_1: (x-\alpha)^2 + (y-\beta)^2 = r_1^2$ and $C_2: (x-8)^2 + (y-\frac{15}{2})^2 = r_2^2$ 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 $(\alpha+\beta) + 4(r_1^2 + r_2^2)$ equals

For the four circles $M, N, O$ and $P$,the following four equations are given:
Circle $M: x^2 + y^2 = 1$
Circle $N: x^2 + y^2 - 2x = 0$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$
Circle $P: x^2 + y^2 - 2y = 0$
If the centre of circle $M$ is joined with the centre of circle $N$,the centre of circle $N$ is joined with the centre of circle $O$,the centre of circle $O$ is joined with the centre of circle $P$,and lastly,the centre of circle $P$ is joined with the centre of circle $M$,then these lines form the sides of a:

If the angle between the circles $x^2+y^2+4x-5=0$ and $x^2+y^2+2\lambda y-4=0$ is $\frac{\pi}{3}$,then $\lambda=$

The area of the circle centered at $(1, 2)$ and passing through $(4, 6)$ is

The angle of intersection of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the circle $x^2 + y^2 = ab$ is

The centres of two circles $C_1$ and $C_2$ each of unit radius are at a distance of $6$ units from each other. Let $P$ be the midpoint of the line segment joining the centres of $C_1$ and $C_2$ and $C$ be a circle touching circles $C_1$ and $C_2$ externally. If a common tangent to $C_1$ and $C$ passing through $P$ is also a common tangent to $C_2$ and $C$,then the radius of the circle $C$ is