For what value of $k$ do the circles $x^2 + y^2 + 5x + 3y + 7 = 0$ and $x^2 + y^2 - 8x + 6y + k = 0$ intersect orthogonally?

Let the set of all values of $r$, for which the circles $(x+1)^{2}+(y+4)^{2}=r^{2}$ and $x^{2}+y^{2}-4x-2y-4=0$ intersect at two distinct points, be the interval $(\alpha, \beta)$. Then $\alpha\beta$ is equal to

If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is:

The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve

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 -

If the circle passing through $(3,5), (5,5)$ and $(3,-3)$ cuts the circle $x^2+y^2+2x+2fy=0$ orthogonally, then $f=$