The equation of the line passing through the points of intersection of the circles $3x^2 + 3y^2 - 2x + 12y - 9 = 0$ and $x^2 + y^2 + 6x + 2y - 15 = 0$ is

The circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2+2x+4y-11=0$:

The area of the triangle formed by the tangents from the point $(h, k)$ to the circle $x^2 + y^2 = a^2$ and the line joining their points of contact is

Consider two circles $C_1: x^2+y^2=25$ and $C_2: (x-\alpha)^2+y^2=16$,where $\alpha \in (5, 9)$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_1$ and $C_2$ be $\sin^{-1}\left(\frac{\sqrt{63}}{8}\right)$. If the length of the common chord of $C_1$ and $C_2$ is $\beta$,then the value of $(\alpha \beta)^2$ equals:

The area of the triangle (in sq. units) formed by the tangents drawn from $P(4,4)$ to the circle $S \equiv x^2+y^2-2x-2y-7=0$ and the chord of contact of $P$ with respect to $S=0$ is

The radius of the circle,having centre at $(2, 1)$ and one of its chords as a diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$,is