If the line $ax + y = c$ touches both the curves $x^2 + y^2 = 1$ and $y^2 = 4\sqrt{2}x$,then $|c|$ is equal to

$A$ line drawn through the point $P(4, 7)$ cuts the circle $x^2 + y^2 = 9$ at the points $A$ and $B$. Then $PA \cdot PB$ is equal to

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 . . . . . . .

$A$ circle is inscribed in an equilateral triangle of side $a$. The area of any square inscribed in the 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
$(A)$ $(x+y)^2=3xy$ $(B)$ $x^{2/3}+y^{2/3}=2^{4/3}$ $(C)$ $x^2+y^2=2xy$ $(D)$ $x^2+y^2=x^2y^2$

Let $C_i \equiv x^2 + y^2 = i^2$ for $i = 1, 2, 3$ be three circles. There are $4i$ points on the circumference of each circle $C_i$. If no three of all the points on the three circles are collinear,then the number of triangles that can be formed using these points whose circumcentre does not lie on the origin is: