The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

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:

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

Consider the three circles: $S_{1} \equiv x^{2}+y^{2}-6x-6y+4=0$,$S_{2} \equiv x^{2}+y^{2}-2x-4y+3=0$,and $S_{3} \equiv x^{2}+y^{2}+2kx+2y+1=0$. If the radical centre of these three circles exists,then which of the following cannot be the value of $k$?

If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

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 $\dots\dots$