The vertex of the parabola $(y - 1)^2 = 8(x - 1)$ is at the centre of a circle and the parabola cuts that circle at the ends of its latus rectum. Then the equation of that circle is

The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

The radius of the circle having its centre at $(0, 3)$ and passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

Given the circle $C$ with the equation $x^2+y^2-2x+10y-38=0$. Match the List-$I$ with the List-$II$ given below concerning $C$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

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: