If one of the diameters of the circle $x^{2}+y^{2}-2 x-6 y+6=0$ is a chord of another circle $'C'$, whose center is at $(2,1),$ then its radius is..........

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 :

Consider the circles ${x^2} + {(y - 1)^2} = $ $9,{(x - 1)^2} + {y^2} = 25$. They are such that

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:

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 line $L$ passes through the points of intersection of the circles ${x^2} + {y^2} = 25$ and ${x^2} + {y^2} - 8x + 7 = 0$. The length of perpendicular from centre of second circle onto the line $L$, is