Consider the two curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Then,

If $(x, y)$ is a variable point on the curve $x^2 + y^2 - 2x - 2y - 2 = 0$,then the minimum value of the expression $\frac{8}{(x - 1)^2} - \frac{(y - 1)^2}{4}$ is equal to

$A$ circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has its centre at the point $(2, 5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$,then $3 \beta - 2 \alpha$ is equal to:

If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$,then $r$ is equal to:

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 $L_1$ 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 $L_1$?