The equation of the line perpendicular to the radical axis of two circles $x^2+y^2-5x+6y+12=0$ and $x^2+y^2+6x-4y-14=0$,and passing through $(1,1)$ is:

$A$ 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 $(-8,1)$

The radical centre of three circles described on the three sides of a triangle as diameters is

If $(-1, -1)$ is the radical centre of the circles $x^2 + y^2 + 2gx - 4y + 4 = 0$,$x^2 + y^2 + 6x + 2fy + 12 = 0$,and $x^2 + y^2 + 10y + 20 = 0$,then $g - f = $

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

If a circle passing through the point $(1,1)$ cuts the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-4y+3=0$ orthogonally,then the center of that circle is: