If $x^2+y^2-6x-8y+12=0$ and $x^2+y^2-4x+6y+k=0$ cut orthogonally,then $k=$

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

The equation of the circle which passes through the point $(2a, 0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle $x^2 + y^2 = a^2$ is:

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

The co-axial system of circles given by ${x^2} + {y^2} + 2gx + c = 0$ for $c < 0$ represents

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 .............