If a circle $S$ with radius $5$ touches the circle $x^2+y^2-6x-4y-12=0$ at $(-1,-1)$,then the length of the tangent from the centre of the circle $S$ to the given circle is

The straight 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 point$(s)$ in $S$ lying inside the smaller part is

If the shortest distance from $(2,-14)$ to the circle $x^2+y^2+6x+4y-12=0$ is $d$ and the length of the tangent drawn from the same point to the circle is $l$,then $\sqrt{d+l}=$

Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $\frac{\pi}{2}$ at the centre is

The total number of common tangents of $x^{2}+y^{2}-6x-8y+9=0$ and $x^{2}+y^{2}=1$ is

The number of common tangents to two circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 8x + 12 = 0$ is