A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then

$(A)$ radius of $S$ is $8$

$(B)$ radius of $S$ is $7$

$(C)$ centre of $S$ is $(-7,1)$

$(D)$ centre of $S$ is $(-8,1)$

The number of common tangents to the circles ${x^2} + {y^2} – 4x – 6y – 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ 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:

The value of $\lambda $, for which the circle ${x^2} + {y^2} + 2\lambda x + 6y + 1 = 0$, intersects the circle ${x^2} + {y^2} + 4x + 2y = 0$ orthogonally is

Circles ${x^2} + {y^2} – 2x – 4y = 0$ and ${x^2} + {y^2} – 8y – 4 = 0$