Two given circles ${x^2} + {y^2} + ax + by + c = 0$ and ${x^2} + {y^2} + dx + ey + f = 0$ will intersect each other orthogonally, only when
$a + b + c = d + e + f$
$ad + be = c + f$
$ad + be = 2c + 2f$
$2ad + 2be = c + f$
The equation of the circle which passing through the point $(2a,\,0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle ${x^2} + {y^2} = {a^2},$ will be
If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then
One of the limit point of the coaxial system of circles containing ${x^2} + {y^2} - 6x - 6y + 4 = 0$, ${x^2} + {y^2} - 2x$ $ - 4y + 3 = 0$ is
Radical axis of the circles $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ and ${x^2} + {y^2} - 3x - 4y + 5 = 0$ is
The range of values of $'a'$ such that the angle $\theta$ between the pair of tangents drawn from the point $(a, 0)$ to the circle $x^2 + y^2 = 1$ satisfies $\frac{\pi }{2} < \theta < \pi$ is :