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

If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point

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

Circles ${(x + a)^2} + {(y + b)^2} = {a^2}$ and ${(x + \alpha )^2}$ $ + {(y + \beta )^2} = $ ${\beta ^2}$ cut orthogonally, if