Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $\mathrm{I}$ with center at the point $A=(4,1)$, where $1<\mathrm{r}<3$. Two distinct common tangents $P Q$ and $S T$ of $C_1$ and $C_2$ are drawn. The tangent $P Q$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $S T$ touches $C_1$ at $S$ and $C_2$ at $T$. Mid points of the line segments $P Q$ and $S T$ are joined to form a line which meets the $x$-axis at a point $B$. If $A B=\sqrt{5}$, then the value of $r^2$ is

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 line $L$ passes through the points of intersection of the circles ${x^2} + {y^2} = 25$ and ${x^2} + {y^2} – 8x + 7 = 0$. The length of perpendicular from centre of second circle onto the line $L$, is