The centre of the circle,which cuts orthogonally each of the three circles $x^2 + y^2 + 2x + 17y + 4 = 0$,$x^2 + y^2 + 7x + 6y + 11 = 0$,and $x^2 + y^2 - x + 22y + 3 = 0$,is

The straight line $x \cos \alpha + y \sin \alpha = p$ cuts the circle $x^2 + y^2 - a^2 = 0$ at $A$ and $B$. Then the equation of the circle having $AB$ as diameter is

The equation of the circle having the chord $x \cos \alpha + y \sin \alpha = p$ of the circle $x^2 + y^2 = a^2$ as its diameter is:

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 $\sqrt{6}$)

The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$ is:

From a point $P$ on the circle $x^2+y^2-4x-6y+9=0$,a pair of tangents $PQ$ and $PR$ are drawn touching the circle $x^2+y^2-4x-6y+12=0$ at $Q$ and $R$. If $C$ is the centre of the concentric circles,then the area of the $\triangle CQR$ (in sq. units) is