If the circles $x^2+y^2-6x-8y-12=0$ and $x^2+y^2-4x+6y+k=0$ are orthogonal to each other,then the value of $k$ is:

If $y = 2x$ is a chord of the circle $x^2 + y^2 - 10x = 0$,then the equation of the circle of which this chord is a diameter is:

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to

In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

If the angle between the circles $x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y+c=0$ is $\frac{\pi}{4}$,then $c=$

Two orthogonal circles are such that the area of one is twice the area of the other. If the radius of the smaller circle is $r$,then the distance between their centers will be -