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

The equation of the family of circles having the same radical axis as the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 2x + 4y = 6$ is:

If $(a, b)$ and $(c, d)$ are the internal and external centres of similitude of the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-6y+8=0$ respectively,then $(a+d)(b+c)=$

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 - 2x - 2y - 2 = 0$ and $x^2 + y^2 - 6x - 6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles,then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $sq. \, units$.

If one of the diameters of the circle,given by the equation $x^2 + y^2 - 4x + 6y - 12 = 0$,is a chord of a circle $S$ whose center is at $(-3, 2)$,then the radius of $S$ 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: