The radius of the circle whose centre lies at $(1, 2)$,while cutting the circle $x^2 + y^2 + 4x + 16y - 30 = 0$ orthogonally,is (in units):

From any point on the circle $x^2 + y^2 = a^2$,tangents are drawn to the circle $x^2 + y^2 = a^2 \sin^2 \alpha$. The angle between them is:

If one of the diameters of the curve $x^{2}+y^{2}-4x-6y+9=0$ is a chord of a circle with centre $(1,1)$,the radius of this circle is

The circles $x^2 + y^2 + 4x + 6y + 3 = 0$ and $2(x^2 + y^2) + 6x + 4y + C = 0$ will cut orthogonally,if $C$ equals

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

If the curves $x^{2}-6x+y^{2}+8=0$ and $x^{2}-8y+y^{2}+16-k=0$ $(k>0)$ touch each other at a point,then the largest value of $k$ is