Choose the correct statement about two circles whose equations are given below:
$x^{2}+y^{2}-10x-10y+41=0$
$x^{2}+y^{2}-22x-10y+137=0$

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. Given that $P(-5, 6)$ is a point such that $PA=PB=10$,find the equation of line $l$.

Let a circle $C$ touch the lines $L_{1}: 4x - 3y + K_{1} = 0$ and $L_{2}: 4x - 3y + K_{2} = 0$,where $K_{1}, K_{2} \in R$. If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(-1, 2)$ and $L_{2}$ at $(3, -6)$,then the equation of the circle $C$ is:

If the area of an equilateral triangle inscribed in the circle $x^2 + y^2 + 10x + 12y + c = 0$ is $27\sqrt{3} \text{ sq. units}$,then $c$ is equal to

The radical axis of two circles and the line joining their centres are:

Let $A(2,3)$,$B(3,-1)$,and $C(-3,2)$ be three points. If the centre of the circle passing through $A$,$B$,and $C$ is $O(h, k)$,then find the value of $2k - 4h$.