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

Let $P(3 \cos \alpha, 2 \sin \alpha)$,$\alpha \neq 0$,be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,$Q$ be a point on the circle $x^2 + y^2 - 14x - 14y + 82 = 0$,and $R$ be a point on the line $x + y = 5$ such that the centroid of the triangle $PQR$ is $(2 + \cos \alpha, 3 + \frac{2}{3} \sin \alpha)$. Then the sum of the ordinates of all possible points $R$ is:

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
$(A)$ $SP=2\sqrt{5}$
$(B)$ $SQ:QP=(\sqrt{5}+1):2$
$(C)$ the $x$-intercept of the normal to the parabola at $P$ is $6$
$(D)$ the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

If the shortest distance between the parabola $y^2=4x$ and the circle $x^2+y^2-4x-16y+64=0$ is $d$,then $d^2$ is equal to:

For the circle $C$ with the equation $x^2+y^2-16x-12y+64=0$,match the List-$I$ with the List-$II$ given below.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

The correct match is:

The area (in $sq. units$) of the smaller of the two circles that touch the parabola $y^2 = 4x$ at the point $(1, 2)$ and the $x$-axis is