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 $a+2b=$

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$. Then

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$ is:

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