The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to

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 equation of a circle passing through origin and co-axial to circles ${x^2} + {y^2} = {a^2}$ and ${x^2} + {y^2} + 2ax = 2{a^2},$ 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 circles ${x^2} + {y^2} - 10x + 16 = 0$ and ${x^2} + {y^2} = {r^2}$ intersect each other in two distinct points, if

If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x$ $-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}$ $+(y-2 \sqrt{2})^{2}=r^{2}$, then the value of $r^{2}$ is equal to