The point of contact of the given circles $x^2 + y^2 - 6x - 6y + 10 = 0$ and $x^2 + y^2 = 2$ is

Number of circles intersecting $x^2+y^2=4$,$x^2+y^2-2x-3=0$ and $x^2+y^2-2y-3=0$ orthogonally is

The lengths of tangents from a fixed point to three circles of a coaxial system are $t_1, t_2, t_3$. If $P, Q,$ and $R$ are the centers of these circles,then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

$A$ circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$

If two circles which pass through the points $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ cut orthogonally,then:

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is: