The circle ${x^2 + y^2 + 2gx + 2fy + c = 0}$ bisects the circumference of the circle ${x^2 + y^2 + 2g'x + 2f'y + c' = 0}$,if

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

Find the equation of the circle passing through the point $(1, 1)$ and the intersection points of the circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$.

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$

Suppose $A(2,3)$ and $B$ are the points of intersection of two circles. The points $P$ lying on one circle and $Q$ lying on the other circle are such that $BP$ and $BQ$ constitute the diameters of the circles. If the slopes of the radical axis and $PQ$ are $3/4$ and $a/b$ respectively,then the value of $3a+4b$ is

The equation of a circle passing through the points of intersection of the circles $x^2 + y^2 + 13x - 3y = 0$ and $2x^2 + 2y^2 + 4x - 7y - 25 = 0$ and the point $(1, 1)$ is