If the circle $x^2+y^2-6x-12y+1=0$ cuts another circle $C$ orthogonally and the centre of the circle $C$ is $(-4, 2)$,then its radius is

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2px=0$ and $x^2+y^2-2qy=0$ and having its centre on the line $\frac{x}{p}-\frac{y}{q}=2$ is:

If $2 x^{2}+2 y^{2}+4 x+5 y+1=0$ and $3 x^{2}+3 y^{2}+6 x-7 y+3 k=0$ are orthogonal,then the value of $k$ is

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):