For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 - 2y = 0$,there is/are:

The number of common tangents to the circles $x^2 + y^2 - 2x - 1 = 0$ and $x^2 + y^2 - 2y - 7 = 0$ is:

Let the line $x - y = 4$ intersect the circle $C : (x - 4)^2 + (y + 3)^2 = 9$ at the points $Q$ and $R$. If $P(\alpha, \beta)$ is a point on $C$ such that $PQ = PR$, then $(6\alpha + 8\beta)^2$ is equal to . . . . . .

Consider the circles ${x^2} + {(y - 1)^2} = 9$ and ${(x - 1)^2} + {y^2} = 25$. They are such that:

$A$ square is inscribed in the circle $x^2+y^2-2x+4y-93=0$ with its sides parallel to the coordinate axes. Which of the following can be one of the vertices of the square?

$A$ 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 $3$.