If the common chord of the circles $x^2+y^2-2x+2y+1=0$ and $x^2+y^2-2x-2y-2=0$ is the diameter of a circle $S$,then the centre of the circle $S$ is

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

$L_1$ 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

The length of the common chord of the circles ${x^2} + {y^2} + 5x + 7y + 9 = 0$ and ${x^2} + {y^2} + 7x + 5y + 9 = 0$ is

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 chord of contact of tangents from a point on the circle $x^2+y^2=r_1^2$ to the circle $x^2+y^2=r_2^2$ touches the circle $x^2+y^2=r_3^2$,then $r_1, r_2, r_3$ are in: