The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ intersect at two distinct points, is ${R}-[a, b]$, then the point $(8 a+12,16 b-20)$ lies on the curve:

If the two circles, $x^2 + y^2 + 2 g_1x + 2 f_1y = 0\, \& \,x^2 + y^2 + 2 g_2x + 2 f_2y = 0$ touch each then:

Suppose $S_1$ and $S_2$ are two unequal circles, $A B$ and $C D$ are the direct common tangents to these circles. A transverse common tangent $P Q$ cuts $A B$ in $R$ and $C D$ in $S$. If $A B=10$, then $R S$ is

The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is 

If the two circles $2{x^2} + 2{y^2} - 3x + 6y + k = 0$ and ${x^2} + {y^2} - 4x + 10y + 16 = 0$ cut orthogonally, then the value of $k$ is