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:

The gradient of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ is

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ is

If one of the diameters of the circle,given by the equation $x^2+y^2-4x+6y-12=0$,is a chord of a circle,$S$,whose centre is at $(-3,2)$,then the length of the radius of $S$ is . . . . . . units.

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is:

If $y = 2x$ is a chord of the circle $x^{2} + y^{2} = 10x$,then find the equation of the circle having this chord as its diameter.