$A$ variable line $ax + by + c = 0$,where $a, b, c$ are in $A.P.$,is normal to a circle $(x - \alpha)^2 + (y - \beta)^2 = \gamma$,which is orthogonal to the circle $x^2 + y^2 - 4x - 4y - 1 = 0$. The value of $\alpha + \beta + \gamma$ 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
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2x+2y=0$,then:

If $y = m_{1}x + c_{1}$ and $y = m_{2}x + c_{2}$ with $m_{1} \neq m_{2}$ are two common tangents of the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = x$,then the value of $8|m_{1}m_{2}|$ is equal to

If from any point on the circle $x^2+y^2+2gx+2fy+c=0$,tangents are drawn to the circle $x^2+y^2+2gx+2fy+c \sin^2 \alpha + (g^2+f^2) \cos^2 \alpha = 0$,where $0 < \alpha < \frac{\pi}{2}$,then the angle between those tangents is

Two circles each of radius $5$ units touch each other at $(1,2)$ and $4x+3y=10$ is their common tangent. The equation of that circle among the two given circles,such that some portion of it lies in every quadrant is