The number of real circles cutting orthogonally the circle $x^{2}+y^{2}+2x-2y+7=0$ is

If the line $x \cos \theta + y \sin \theta = 2$ is the equation of a transverse common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6 \sqrt{3} x - 6y + 20 = 0$,then the value of $\theta$ is:

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

The angle of intersection between the curves $y^2+x^2=a^2 \sqrt{2}$ and $x^2-y^2=a^2$ is

$A(x_1, y_1)$ is the internal centre of similitude and $B(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are $P(\alpha, \beta)$ and $Q(\gamma, \delta)$ respectively. If $PA=3, AB=5, QB=2$,then the ratio of the radii of the two circles is:

The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circle $S=0$ is $t$,then the point $(2, 3)$ with respect to the circle $S^{\prime}=0$ having centre at $(p, t^2)$ and passing through the origin: