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 range of values of $'a'$ such that the angle $\theta$ between the pair of tangents drawn from the point $(a, 0)$ to the circle $x^2 + y^2 = 1$ satisfies $\frac{\pi }{2} < \theta < \pi$ is :

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y – 24 = 0$ also passes through the point

Give the number of common tangents to circle ${x^2} + {y^2} + 2x + 8y – 23 = 0$ and ${x^2} + {y^2} – 4x – 10y + 9 = 0$

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 – 4x – 6y – 21 = 0$ and $3x + 4y + 5 = 0$ is given by