Find the equation of the chord of contact of the tangents drawn from the point $(2, -3)$ to the circle $x^2 + y^2 + 4x - 6y - 12 = 0$.

Tangents are drawn from the point $(-1, -4)$ to the circle $x^2 + y^2 - 2x + 4y + 1 = 0$. The length of the corresponding chord of contact 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 $\alpha$)

The locus of the point of intersection of the tangents at the extremities of a chord of the circle $x^2 + y^2 = a^2$ which touches the circle $x^2 + y^2 = 2ax$ is

$L_1$ and $L_2$ are two common tangents to two circles. If $L_1$ touches the two circles at $A(1, 1)$ and $B(0, 1)$ and $L_2$ touches the two circles at $C\left(\frac{3}{5}, \frac{4}{5}\right)$ and $D\left(-\frac{1}{5}, \frac{7}{5}\right)$,then the equation of the radical axis of the two circles is

Given the circles $x^2 + y^2 - 4x - 5 = 0$ and $x^2 + y^2 + 6x - 2y + 6 = 0$. Let $P$ be a point $(\alpha, \beta)$ such that the lengths of the tangents from $P$ to both circles are equal. Then: