The equation of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $2{x^2} + 2{y^2} - 10x - 12y + 12 = 0$ is:

The equation of a circle,which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$,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

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A=(4,1)$,where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Midpoints of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB=\sqrt{5}$,then the value of $r^2$ is

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is: