The radical centre of the circles $x^2+y^2+3x+2y+1=0$,$x^2+y^2-x+6y+5=0$ and $x^2+y^2+5x-8y+15=0$ is

The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is

The locus of the centres of the circles which touch externally the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 4ax = 0$ will be

The equation of the circle passing through the origin and cutting the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ orthogonally is

If the coordinates of the point of contact of the circles $x^2+y^2-4x+8y+4=0$ and $x^2+y^2+2x=0$ are $(a, b)$,then $a+2b=$

Statement $(A) :$ If two circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other,then $f'g = fg'$.
Reason $(R) :$ Two circles touch each other if the line joining their centers is perpendicular to all possible common tangents.