$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

$A$ circle is drawn in a sector of a larger circle of radius $r$,as shown in the figure. The smaller circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is

If the length of the chord $2x+3y+k=0$ of the circle $x^2+y^2-6x-8y+9=0$ is $2\sqrt{3}$,then one of the values of $k$ is

Let $ABCD$ be a square of side length $1$,and $\Gamma$ be a circle passing through $B$ and $C$,and touching $AD$. The radius of $\Gamma$ is

Suppose the tangents drawn to the circle $x^2+y^2-6x-4y-11=0$ from $P(1,8)$ touch the circle at $A$ and $B$. Then the centre of the circle passing through $P, A$ and $B$ is

The line $2x - 3y = 1$ divides the circular region $x^2 + y^2 \leq 6$ into two parts. If $S = \left\{ \left(2, \frac{3}{4}\right), \left(\frac{5}{2}, \frac{3}{4}\right), \left(\frac{1}{4}, -\frac{1}{4}\right), \left(\frac{1}{8}, \frac{1}{4}\right) \right\}$,then the number of points in the set $S$ that lie inside the smaller part is: