The internal centre of similitude of the two circles $x^2+y^2-4x-6y+12=0$ and $x^2+y^2+4x-2y-4=0$ is

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

Find the value of $m+n$,if the circumference of the circle $x^2+y^2+8x+8y-m=0$ is bisected by the circle $x^2+y^2-2x+4y+n=0$.

The equation of the circle which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$ is

If the circles $x^{2}+y^{2}+6x+8y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y = k+6\sqrt{3}+8\sqrt{6}$ with $k>0$ touch internally at the point $P(\alpha, \beta)$,then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1 = x^2 + y^2 + kx - 4y - 1 = 0$ and $S_2 = 3x^2 + 3y^2 - 14x + 23y - 15 = 0$ and passing through the point $(-1, -1)$ is: