If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x$ $-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}$ $+(y-2 \sqrt{2})^{2}=r^{2}$, then the value of $r^{2}$ is equal to

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

Two circle ${x^2} + {y^2} = ax$ and ${x^2} + {y^2} = {c^2}$ touch each other if 

The circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2g'x + 2f'y + c' = 0$, if

If $d$ is the distance between the centres of two circles, ${r_1},{r_2}$ are their radii and $d = {r_1} + {r_2}$, then

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