Find the equation of the radical axis of the two circles $x^2 + y^2 - x + 1 = 0$ and $3(x^2 + y^2) + y - 1 = 0$.

If the circles $x^2+y^2-4x+2fy+1=0$ and $x^2+y^2+2gx-4y-1=0$ cut orthogonally,then $r_1^2+r_2^2-8=$

The radical axis of the co-axial system of circles with limiting points $(1, 2)$ and $(-2, 1)$ is

If a circle has its centre on the line $x-y-1=0$ and passes through the points of intersection of the two circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y-7=0$,then the centre of that circle is

If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is:

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is: