If a circle passes through the points $(0,0), (x,0),$ and $(0,y)$,then the coordinates of its centre are

$A$ 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 $3$.

The line $4x + 3y - 4 = 0$ divides the circumference of a circle in the ratio $1:2$. If $C(5, 3)$ is the center of that circle,then the equation of the circle is

If $h, k, p, q \neq 0$ and the circles $x^2+y^2+2hx+2ky=0$ and $x^2+y^2+2px+2qy=0$ touch each other at the origin,then $hq-pk-\frac{hq}{pk}$ is equal to

The equation of the circle which passes through $(1, 0)$ and $(0, 1)$ and has its radius as small as possible,is

In a circle with center $O$,suppose $A, P, B$ are three points on its circumference such that $P$ is the mid-point of the minor arc $AB$. Suppose when $\angle AOB = \theta$,$\frac{\text{area}(\triangle AOB)}{\text{area}(\triangle APB)} = \sqrt{5} + 2$. If $\angle AOB$ is doubled to $2\theta$,then the ratio $\frac{\text{area}(\triangle AOB)}{\text{area}(\triangle APB)}$ is