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:

Three concentric circles,of which the biggest is $x^2 + y^2 = 1$,have their radii in $A.P.$ If the line $y = x + 1$ cuts all the circles in real and distinct points,the interval in which the common difference $d$ of the $A.P.$ will lie is:

If the point of intersection of the pair of the transverse common tangents and that of the pair of direct common tangents drawn to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$ are $T$ and $D$ respectively,then the centre of the circle having $TD$ as diameter is

Let $a$ and $b$ be non-zero real numbers. Then,the equation $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$ represents

Let a circle $S = 0$ touch both the circles $x^2 + y^2 = 400$ and $x^2 + y^2 - 10x - 24y + 120 = 0$ externally and also touch the $x$-axis. The radius of the circle $S = 0$ is

The circle $x^2+y^2-8x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.
$1.$ Equation of a common tangent with positive slope to the circle as well as to the hyperbola is:
$(A) 2x-\sqrt{5}y-20=0$
$(B) 2x-\sqrt{5}y+4=0$
$(C) 3x-4y+8=0$
$(D) 4x-3y+4=0$
$2.$ Equation of the circle with $AB$ as its diameter is:
$(A) x^2+y^2-12x+24=0$
$(B) x^2+y^2+12x+24=0$
$(C) x^2+y^2+24x-12=0$
$(D) x^2+y^2-24x-12=0$