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:

$A$ 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
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

Consider the curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Assertion $(A)$: The common tangents to the curves $C_1$ and $C_2$ are orthogonal. Reason $(R)$: $x-y+1=0$ and $x+y+1=0$ are the common tangents to the curves $C_1$ and $C_2$.

Let circle $C$ be the image of $x^2+y^2-2x+4y-4=0$ in the line $2x-3y+5=0$. Let $A$ be the point on $C$ such that $OA$ is parallel to the $x$-axis and $A$ lies on the right hand side of the centre $O$ of $C$. If $B(\alpha, \beta)$,with $\beta < 4$,lies on $C$ such that the length of the arc $AB$ is $(1/6)^{\text{th}}$ of the perimeter of $C$,then $\beta - \sqrt{3}\alpha$ is equal to

Three concentric circles,of which the biggest is $x^2 + y^2 = 1$,have their radii in $A.P.$ If the line $y = x + 1$ cuts all the circles in real and distinct points,the interval in which the common difference $d$ of the $A.P.$ will lie is: