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:

For different real non-zero numbers $x_1, x_2, x_3$ and $x_4$,suppose the points $(x_1, \frac{1}{x_1}), (x_2, \frac{1}{x_2}), (x_3, \frac{1}{x_3})$ and $(x_4, \frac{1}{x_4})$ lie on the boundary of a circle of radius $4$. Then,the value of $x_1 x_2 x_3 x_4$ is

$A$ tangent $PT$ is drawn to the circle $x^2+y^2=4$ at the point $P(\sqrt{3}, 1)$. $A$ straight line $L$,perpendicular to $PT$,is a tangent to the circle $(x-3)^2+y^2=1$.
$1.$ $A$ common tangent of the two circles is
$(A)$ $x=4$ $(B)$ $y=2$ $(C)$ $x+\sqrt{3} y=4$ $(D)$ $x+2 \sqrt{2} y=6$
$2.$ $A$ possible equation of $L$ is
$(A)$ $x-\sqrt{3} y=1$ $(B)$ $x+\sqrt{3} y=1$ $(C)$ $x-\sqrt{3} y=-1$ $(D)$ $x+\sqrt{3} y=5$

Let $a$ and $b$ be two non-zero real numbers. The equation $(ax^2 + by^2 + c)(x^2 - 5xy + 6y^2) = 0$ represents:

Let $L_1$ be a straight line passing through the origin and $L_2$ be the straight 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 can represent $L_1$?

If $\theta$ is the angle subtended at $P(x_1, y_1)$ by the circle $S \equiv x^2 + y^2 + 2gx + 2fy + c = 0$,then