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 $\dots\dots$

From a point $P$ on the circle $x^2+y^2-4x-6y+9=0$,a pair of tangents $PQ$ and $PR$ are drawn touching the circle $x^2+y^2-4x-6y+12=0$ at $Q$ and $R$. If $C$ is the centre of the concentric circles,then the area of the $\triangle CQR$ (in sq. units) is

Let $A(1, 2)$ be the centre and $3$ be the radius of a circle $S$. Let $B(-1, -1)$ be the centre and $r$ be the radius of another circle $S^{\prime}$. If $\frac{\pi}{3}$ is the angle between the circles $S$ and $S^{\prime}$,then the number of possible values of $r$ is

If $C_1$ and $C_2$ are the centres of similitude with respect to the circles $x^2+y^2+6x+8y+24=0$ and $x^2+y^2-6x-8y+9=0$,then $C_1C_2=$

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :