The equation of the circle which cuts all the three circles $4(x-1)^2+4(y-1)^2=1$,$4(x+1)^2+4(y-1)^2=1$,and $4(x+1)^2+4(y+1)^2=1$ orthogonally is

The equation of the radical axis of the pair of circles $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$ is

Let $a=1+i$ and $z=x+iy$. If the curve $z\bar{z}+az+\bar{a}\bar{z}-4=0$ is cut by the straight line $(z+\bar{z})-i(z-\bar{z})+2=0$ at two points $A$ and $B$, then the equation of the circle passing through the origin, $A$ and $B$ 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:

The equation of the tangent at the point $(0, 3)$ on the circle which cuts the circles $x^2 + y^2 - 2 x + 6 y = 0$,$x^2 + y^2 - 4 x - 2 y + 6 = 0$ and $x^2 + y^2 - 12 x + 2 y + 3 = 0$ orthogonally is

Find the equation of the circle which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$,$x^2+y^2+5x-5y+9=0$,and $x^2+y^2+7x-9y+29=0$.