The length of the common chord of the circles $x^2 + y^2 + 2x + 4y - 20 = 0$ and $x^2 + y^2 + 6x - 8y + 10 = 0$ is

The equation of the common tangent to the circles $x^2+y^2-4x+10y+20=0$ and $x^2+y^2+8x-6y-24=0$ is

If the chord $L \equiv y-mx-1=0$ of the circle $S \equiv x^2+y^2-1=0$ touches the circle $S_1 \equiv x^2+y^2-4x+1=0$,then the possible points for which $L=0$ is a chord of contact of $S=0$ are

$A$ 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 $\sqrt{3}$)

Tangents are drawn from the point $(4, 3)$ to the circle $x^2 + y^2 = 9$. The area of the triangle formed by them and the line joining their points of contact is

If the circles $x^2 + y^2 + 2ax + cy + a = 0$ and $x^2 + y^2 - 3ax + dy - 1 = 0$ intersect in two distinct points $P$ and $Q$,then the line $5x + by - a = 0$ passes through $P$ and $Q$ for