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

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
$(A)$ $(x+y)^2=3xy$ $(B)$ $x^{2/3}+y^{2/3}=2^{4/3}$ $(C)$ $x^2+y^2=2xy$ $(D)$ $x^2+y^2=x^2y^2$

If the distances from the origin to the centres of the three circles $x^2 + y^2 - 2\lambda_i x = c^2$ for $i = 1, 2, 3$ are in $G.P.$,then the lengths of the tangents drawn to them from any point on the circle $x^2 + y^2 = c^2$ are in

$A$ parabola $y = ax^2 + bx + c$ crosses the $x$-axis at $(\alpha, 0)$ and $(\beta, 0)$,both to the right of the origin. $A$ circle also passes through these two points. The length of a tangent from the origin to the circle 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:

If the curves $ax^2+by^2=1$ and $cx^2+dy^2=1$ intersect orthogonally,then $\frac{b-a}{d-c}=$