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 mirror image of a circle $c_{1}: x^{2}+y^{2}-2 x-$ $6 y+\alpha=0$ in line $y=x+1$ be $c_{2}: 5 x^{2}+5 y^{2}+10 g x$ $+10 f y +38=0$. If $r$ is the radius of circle $c _{2}$, then $\alpha+6 r^{2}$ is equal to$.....$

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

If one common tangent of the two circles $x^2 + y^2 = 4$ and ${x^2} + {\left( {y - 3} \right)^2} = \lambda ,\lambda  > 0$ passes through the point $\left( {\sqrt 3 ,1} \right)$, then possible value of  $\lambda$ is

The lengths of tangents from a fixed point to three circles of coaxial system are ${t_1},{t_2},{t_3}$ and if $P, Q$ and $R$ be the centres, then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

If $y = 2x$ is a chord of the circle ${x^2} + {y^2} - 10x = 0$, then the equation of the circle of which this chord is a diameter, is