The equation of the circle whose radius is $3$ and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point $(-1,-1)$ is

Let $A(2, 3), B(4, 5)$ be two points and let $C = (x, y)$ be a point such that $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$. If the area of $\Delta ABC = \sqrt{2}$ sq. unit,then the maximum number of positions of $C$ in the $xy$ plane 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 $3$.

The triangle $PQR$ is inscribed in the circle $x^2+y^2=25$. If $Q=(3,4)$ and $R=(-4,3)$,then $\angle QPR=$

$A$ circle $C$ touches the $X$-axis and makes an intercept of length $2$ units on the $Y$-axis. If the centre of this circle lies on the line $y=x+1$,then which of the following is a circle passing through the centre of the circle $C$?

If $(x_i, y_i)$ are the vertices of an equilateral triangle $ABC$ such that $(x_1 - 2)^2 + (y_1 - 3)^2 = (x_2 - 2)^2 + (y_2 - 3)^2 = (x_3 - 2)^2 + (y_3 - 3)^2$,then find the value of $2(x_1 + x_2 + x_3) + 3(y_1 + y_2 + y_3)$.