Choose the correct statement about two circles whose equations are given below

$x^{2}+y^{2}-10 x-10 y+41=0$

$x^{2}+y^{2}-22 x-10 y+137=0$

The equation of the circle which touches the circle ${x^2} + {y^2} - 6x + 6y + 17 = 0$ externally and to which the lines ${x^2} - 3xy - 3x + 9y = 0$ are normals, is

Let

$A=\left\{(x, y) \in R \times R \mid 2 x^{2}+2 y^{2}-2 x-2 y=1\right\}$

$B=\left\{(x, y) \in R \times R \mid 4 x^{2}+4 y^{2}-16 y+7=0\right\} \text { and }$

$C=\left\{(x, y) \in R \times R \mid x^{2}+y^{2}-4 x-2 y+5 \leq r^{2}\right\}$

Then the minimum value of $|r|$ such that $A \cup B \subseteq C$ is equal to:

The value of k so that ${x^2} + {y^2} + kx + 4y + 2 = 0$ and $2({x^2} + {y^2}) - 4x - 3y + k = 0$ cut orthogonally is

The number of common tangents to the circles ${x^2} + {y^2} = 1$and ${x^2} + {y^2} - 4x + 3 = 0$ is

Let $S = 0$ is the locus of centre of a variable circle which intersect the circle $x^2 + y^2 -4x -6y = 0$ orthogonally at $(4, 6)$ . If $P$ is a variable point of $S = 0$ , then least value of $OP$ is (where $O$ is origin)