Equation of radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$, $2{x^2} + 2{y^2} - 10x$$ - 12y + 12 = 0$ is

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 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-

A circle ${C_1}$ of radius $2$ touches both $x$ - axis and $y$ - axis. Another circle ${C_2}$ whose radius is greater than $2$ touches circle ${C_1}$ and both the axes. Then the radius of circle ${C_2}$ is

The centre of the circle, which cuts orthogonally each of the three circles ${x^2} + {y^2} + 2x + 17y + 4 = 0,$ ${x^2} + {y^2} + 7x + 6y + 11 = 0,$ ${x^2} + {y^2} - x + 22y + 3 = 0$ is

Radius of circle touching $y-$axis at point $P(0,2)$ and circle $x^2 + y^2 = 16$ internally-