If the circles $(x+1)^2+(y+2)^2=r^2$ and $x^2+y^2-4 x-4 y+4=0$ intersect at exactly two distinct points, then

The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing  through the points of intersection of the two circles $x^2 + y^2 + 5x -8y + 1 =0$ and $x^2 + y^2-3x + 7y -25 = 0$ is-

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$, is

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 

The value of $\lambda $, for which the circle ${x^2} + {y^2} + 2\lambda x + 6y + 1 = 0$, intersects the circle ${x^2} + {y^2} + 4x + 2y = 0$ 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$.....$