If the circles ${x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}$ and ${(x - 4)^2} + {(y - 7)^2} = 36$ intersect at two distinct points, then

The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be 

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be

The circle passing through point of intersection of the circle $S = 0$ and the line $P = 0$ is

Two circle ${x^2} + {y^2} = ax$ and ${x^2} + {y^2} = {c^2}$ touch each other if 

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: