The equation of a circle passing through the vertex and the extremities of the latus rectum of the parabola ${y^2 = 8x}$ is

Let $C$ be the circle $x^2+(y-1)^2=2$. Let $E_1$ and $E_2$ be two ellipses whose centers lie at the origin and whose major axes lie on the $x$-axis and $y$-axis,respectively. Let the straight line $x+y=3$ touch the curves $C$,$E_1$,and $E_2$ at $P(x_1, y_1)$,$Q(x_2, y_2)$,and $R(x_3, y_3)$,respectively. Given that $P$ is the midpoint of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$,the value of $9(x_1y_1 + x_2y_2 + x_3y_3)$ is equal to . . . . . . .

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:

$A$ focal chord to $y^2 = 16x$ is a tangent to $(x - 6)^2 + y^2 = 2$. Then the possible values of the slope of this chord are:

If the point $(2, \lambda)$ lies inside the circles $x^2+y^2=13$ and $x^2+y^2+x-2y=14$,then $\lambda$ lies in the set

Consider the curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Assertion $(A)$: The common tangents to the curves $C_1$ and $C_2$ are orthogonal. Reason $(R)$: $x-y+1=0$ and $x+y+1=0$ are the common tangents to the curves $C_1$ and $C_2$.