Find the equation of the circle passing through the point $(1, 1)$ and the intersection points of the circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$.

The radius of the circle whose centre lies at $(1, 2)$,while cutting the circle $x^2 + y^2 + 4x + 16y - 30 = 0$ orthogonally,is (in units):

If $T_1 T_1^{\prime}$ and $T_2 T_2^{\prime}$ are the common tangents of the circles $S = x^2 + y^2 - 2x - 4y - 4 = 0$ and $S^{\prime} = x^2 + y^2 + 4x + 4y + 4 = 0$, where $T_1, T_1^{\prime}, T_2, T_2^{\prime}$ are the points of contact, then the distance between $T_1$ and $T_1^{\prime}$ is (in $\sqrt{6}$)

If $x^2+y^2-4x-2y+5=0$ and $x^2+y^2-6x-4y-3=0$ are members of a coaxial system of circles,then the centre of a point circle in the system is

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is