The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ is

If $T_1 T_1^{\prime}$ and $T_2 T_2^{\prime}$ are the common tangents of the circles $S = x^2 + y^2 - 2x - 4y - 4 = 0$ and $S^{\prime} = x^2 + y^2 + 4x + 4y + 4 = 0$, where $T_1, T_1^{\prime}, T_2, T_2^{\prime}$ are the points of contact, then the distance between $T_1$ and $T_1^{\prime}$ is (in $\sqrt{6}$)

$A$ circle $S \equiv x^2+y^2-16=0$ intersects another circle $S^{\prime}=0$ of radius $5$ units such that their common chord is of maximum length. If the slope of that chord is $\frac{3}{4}$,then the centre of such a circle $S^{\prime}=0$ is

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

The length of the tangent drawn from any point on the circle $x^2 + y^2 + 2gx + 2fy + \alpha = 0$ to the circle $x^2 + y^2 + 2gx + 2fy + \beta = 0$ is:

The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is