Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ intersect at two distinct points, is ${R}-[a, b]$, then the point $(8 a+12,16 b-20)$ lies on the curve:

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

Coordinates of the centre of the circle which bisects the circumferences of the circles

$x^2 + y^2 = 1 ; x^2 + y^2 + 2x – 3 = 0$ and $x^2 + y^2 + 2y – 3 = 0$ is

The lengths of tangents from a fixed point to three circles of coaxial system are ${t_1},{t_2},{t_3}$ and if $P, Q$ and $R$ be the centres, then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

If the circles of same radius a and centers at $(2, 3)$ and $(5, 6)$ cut orthogonally, then $a =$