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 the shortest distance from $(a, 0)$,$a > 0$,to the parabola $y^2 = 4x$ be $4$. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola,and having its centre on the axis of the parabola is:

Let $S$ be the circle in the $xy$-plane defined by the equation $x^2+y^2=4$.
$(1)$ Let $E_1, E_2$ and $F_1, F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis,respectively. Let $G_1, G_2$ be the chord of $S$ passing through $P_0$ and having slope $-1$. Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$,the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$,and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then,the points $E_3, F_3$,and $G_3$ lie on the curve
$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $xy=4$
$(2)$ Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then,the mid-point of the line segment $MN$ must lie on the curve
$(A)$ $(x+y)^2=3xy$ $(B)$ $x^{2/3}+y^{2/3}=2^{4/3}$ $(C)$ $x^2+y^2=2xy$ $(D)$ $x^2+y^2=x^2y^2$

Consider two curves $C_1 : y^2 = 2x$ and $C_2 : x^2 + y^2 - 3x + 2 = 0$. Then,

Let $L_1$ be a line passing through the origin and $L_2$ be the line $x + y = 1$. If the intercepts made by the circle $x^{2} + y^{2} - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations represents $L_1$?

The equation of the circle symmetric to the circle $x^2 + y^2 - 2x - 4y + 4 = 0$ about the line $x - y = 3$ is