If one of the two circles $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$ lies within the other,then (where $\alpha_1, \alpha_2 \in R, \alpha_1 \neq \alpha_2$):

The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$ is:

The locus of the centres of the circles which touch externally the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 4ax = 0$ will be

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

If the circles $C_1: x^2+y^2+2x+4y-20=0$ and $C_2: x^2+y^2+6x-8y+9=0$ have $n$ common tangents and the length of the tangent drawn from the centre of similitude to the circle $C_2$ is $l$,then $\frac{l}{n^2} =$