If the circles ${x^2} + {y^2} + 2x + 2ky + 6 = 0$ and ${x^2} + {y^2} + 2ky + k = 0$ intersect orthogonally, then $k$ is
$2$ or $ - \frac{3}{2}$
$-2$ or $\frac{3}{2}$
$2$ or $\frac{3}{2}$
-$2$ or -$\frac{3}{2}$
The set of all real values of $\lambda $ for which exactly two common tangents can be drawn to the circles $x^2 + y^2 - 4x - 4y+ 6\, = 0$ and $x^2 + y^2 - 10x - 10y + \lambda \, = 0$ is the interval:
The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that
A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is
The equation of radical axis of the circles $2{x^2} + 2{y^2} - 7x = 0$ and ${x^2} + {y^2} - 4y - 7 = 0$ is
The centre of the circle, which cuts orthogonally each of the three circles ${x^2} + {y^2} + 2x + 17y + 4 = 0,$ ${x^2} + {y^2} + 7x + 6y + 11 = 0,$ ${x^2} + {y^2} - x + 22y + 3 = 0$ is