Consider a family of circles which are passing through the point $(- 1, 1)$ and are tangent to $x-$ axis. If $(h, k)$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

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

The equation of the circle which passes through the origin, has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally, is

Let the circles $C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$ and $C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$ touch each other externally at the point $(6,6)$. If the point $(6,6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2: 1$, then $(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$ equals

The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals

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