The equation of the radical axis of the circles $2x^2 + 2y^2 - 7x = 0$ and $x^2 + y^2 - 4y - 7 = 0$ is

The equation of the circle which cuts the circles $S_1 \equiv x^2+y^2-4=0$,$S_2 \equiv x^2+y^2-6x-8y+10=0$,and $S_3 \equiv x^2+y^2+2x-4y-2=0$ at the extremities of the diameters of these circles is:

If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

The radical centre of the circles $x^2 + y^2 - 16x + 60 = 0$,$x^2 + y^2 - 12x + 27 = 0$,and $x^2 + y^2 - 12y + 8 = 0$ is