For the two curves $C_1 : y^2 = 4x$ and $C_2 : x^2 + y^2 - 6x + 1 = 0$,which of the following is true?

$A$ circle $S \equiv x^2+y^2+2gx+2fy+6=0$ cuts another circle $x^2+y^2-6x-6y-6=0$ orthogonally. If the angle between the circles $S=0$ and $x^2+y^2+6x+6y+2=0$ is $60^{\circ}$,then the radius of the circle $S=0$ is

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is

The equation of the common tangent touching the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$ above the $X$-axis is

The sum of diameters of the circles that touch $(i)$ the parabola $75x^2 = 64(5y - 3)$ at the point $\left(\frac{8}{5}, \frac{6}{5}\right)$ and $(ii)$ the $y$-axis,is equal to $......$

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$