The diameter of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is a chord of another circle with center $(2, 1)$. Find the radius of this circle.

Let $M = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x^2 + y^2 \leq r^2\}$,where $r > 0$. Consider the geometric progression $a_n = \frac{1}{2^{n-1}}$,$n = 1, 2, 3, \ldots$. Let $S_0 = 0$ and,for $n \geq 1$,let $S_n$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$,let $C_n$ denote the circle with center $(S_{n-1}, 0)$ and radius $a_n$,and $D_n$ denote the circle with center $(S_{n-1}, S_{n-1})$ and radius $a_n$.
$(1)$ Consider $M$ with $r = \frac{1025}{513}$. Let $k$ be the number of all those circles $C_n$ that are inside $M$. Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then
$(A)$ $k + 2l = 22$ $(B)$ $2k + l = 26$ $(C)$ $2k + 3l = 34$ $(D)$ $3k + 2l = 40$
$(2)$ Consider $M$ with $r = \frac{(2^{199}-1)\sqrt{2}}{2^{198}}$. The number of all those circles $D_n$ that are inside $M$ is
$(A)$ $198$ $(B)$ $199$ $(C)$ $200$ $(D)$ $201$

Let $r$ be the radius of the circle,which touches the $x$-axis at point $(a, 0)$,where $a < 0$,and the parabola $y^2 = 9x$ at the point $(4, 6)$. Then $r$ is equal to . . . . . . .

If $(2, a)$ does not lie outside the circles $x^2+y^2=13$ and $x^2+y^2+x-2y=14$,then $a$ lies in

For $n \in N$,let $S_{n} = \{ z \in C : |z - 3 + 2i| = \frac{n}{4} \}$ and $T_{n} = \{ z \in C : |z - 2 + 3i| = \frac{1}{n} \}$. Then the number of elements in the set ${ n \in N : S_{n} \cap T_{n} = \phi }$ is.

The line $x+y=k$ meets the curve $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k$ $(k>1)$ is