The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2+4x+6y+4=0$ is

If the point of contact of the circles $x^2+y^2-6x-4y+9=0$ and $x^2+y^2+2x+2y-7=0$ is $(\alpha, \beta)$, then $7\beta=$ (in $\alpha$)

Find the condition that the pair of lines joining the origin to the points of intersection of the line $y = mx + c$ and the circle $x^{2} + y^{2} = a^{2}$ are at right angles to each other.

If the circles $x^2 + y^2 + 5Kx + 2y + K = 0$ and $2(x^2 + y^2) + 2Kx + 3y - 1 = 0$,$(K \in R)$,intersect at the points $P$ and $Q$,then the line $4x + 5y - K = 0$ passes through $P$ and $Q$ for

Find the distance between the chords of contact of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ with respect to the points $(0, 0)$ and $(g, f)$.

$A$ rhombus is inscribed in the region common to the two circles $x^2+y^2-4x-12=0$ and $x^2+y^2+4x-12=0$. If the line joining the centres of these circles and the common chord of them are the diagonals of this rhombus, then the area (in square units) of the rhombus is (in $\sqrt{3}$)