The equation of a circle which touches both axes and the line $3x - 4y + 8 = 0$ and whose centre lies in the third quadrant is

The number of common tangents to the circles $x^2 + y^2 - 2x + 4y - 4 = 0$ and $x^2 + y^2 - 8x - 4y + 16 = 0$ 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 lengths of the chords intercepted by the circle $x^2 + y^2 + 2gx + 2fy = 0$ from the coordinate axes are $10$ and $24$ respectively,then the radius of the circle is:

Position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is

Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^{2}+y^{2}-2x-6y+6=0$. If these tangents touch the circle at points $A$ and $B$,and if $D$ is a point on the circle such that the lengths of the segments $AB$ and $AD$ are equal,then the area of the triangle $ABD$ is equal to: