If $\theta$ is the angle between the tangents drawn from the point $P(-1, -1)$ to the circle $x^2 + y^2 - 4x - 6y + c = 0$ and $\cos \theta = -\frac{7}{25}$,then the radius of the circle is

The area of the quadrilateral formed by the tangents from the point $(4,5)$ to the circle $x^2+y^2-4x-2y-11=0$,and the pair of radii joining the points of contact of these tangents is

If one end of the diameter of the circle $x^2+y^2-4x-6y+11=0$ is $(3,4)$,then the other end of the diameter is

If one end of a diameter of the circle $3x^{2}+3y^{2}-9x+6y+5=0$ is $(1,2)$, then the other end is

$A$ circle $C$ of radius $1$ is inscribed in an equilateral triangle $PQR$. The points of contact of $C$ with the sides $PQ, QR, RP$ are $D, E, F$,respectively. The line $PQ$ is given by the equation $\sqrt{3}x + y - 6 = 0$ and the point $D$ is $\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$. Further,it is given that the origin and the centre of $C$ are on the same side of the line $PQ$.
$1.$ The equation of circle $C$ is
$(A) (x - 2\sqrt{3})^2 + (y - 1)^2 = 1$
$(B) (x - 2\sqrt{3})^2 + (y + \frac{1}{2})^2 = 1$
$(C) (x - \sqrt{3})^2 + (y + 1)^2 = 1$
$(D) (x - \sqrt{3})^2 + (y - 1)^2 = 1$
$2.$ Points $E$ and $F$ are given by
$(A) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), (\sqrt{3}, 0)$
$(B) \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right), (\sqrt{3}, 0)$
$(C) \left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$(D) \left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right), \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$3.$ Equation of the sides $QR, RP$ are
$(A) y = \frac{2}{\sqrt{3}}x + 1, y = -\frac{2}{\sqrt{3}}x - 1$
$(B) y = \frac{1}{\sqrt{3}}x, y = 0$
$(C) y = \frac{\sqrt{3}}{2}x + 1, y = -\frac{\sqrt{3}}{2}x - 1$
$(D) y = \sqrt{3}x, y = 0$
Give the answer for questions $1, 2$ and $3$.

If the circle $x^2+y^2-4x-8y-5=0$ intersects the line $3x-4y-m=0$ in two distinct points,then the number of integral values of '$m$' is