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:

Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then,which one of the following points lies on the circle $C$?

The equations of the circles which touch the lines $3x - 4y + 1 = 0$ and $4x + 3y - 7 = 0$ and pass through the point $(2, 3)$ are:

If the tangent to the circle $x^2 + y^2 = r^2$ at the point $(a, b)$ meets the coordinate axes at the points $A$ and $B$,and $O$ is the origin,then the area of the triangle $OAB$ is

Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then
$(A)$ $SP=2\sqrt{5}$
$(B)$ $SQ:QP=(\sqrt{5}+1):2$
$(C)$ the $x$-intercept of the normal to the parabola at $P$ is $6$
$(D)$ the slope of the tangent to the circle at $Q$ is $\frac{1}{2}$

Let $y^2 = 4ax$ be a parabola and $x^2 + y^2 + 2bx = 0$ be a circle. If the parabola and the circle touch each other externally,then: