The two circles ${x^2} + {y^2} - 2x + 22y + 5 = 0$ and ${x^2} + {y^2} + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to

In the figure shown, radius of circle $C_1$ be $ r$ and that of $C_2$ be $\frac{r}{2}$ , where $r= \frac {1}{3} PQ,$ then length of $AB$ is (where $P$ and $Q$ being centres of $C_1$ $\&$ $C_2$ respectively)

The circles $x^2 + y^2 + 2x -2y + 1 = 0$ and $x^2 + y^2 -2x -2y + 1 = 0$ touch each  other :-

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

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

The co-axial system of circles given by ${x^2} + {y^2} + 2gx + c = 0$ for $c < 0$ represents