If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

If the line $y = x + 3$ intersects the circle $x^2 + y^2 = a^2$ at points $A$ and $B$,then the equation of the circle with $AB$ as its diameter is:

If the lengths of the tangents drawn from a point $P$ to the three circles $x^2+y^2-4=0$,$x^2+y^2-2x+3y=0$,and $x^2+y^2+7y-18=0$ are equal,then the coordinates of $P$ are

The equation of the circle passing through the points of intersection of the circles $x^2+y^2+4x+6y-12=0$ and $x^2+y^2-6x-4y-12=0$ and cutting the circle $x^2+y^2-4x+4y+8=0$ orthogonally is

The radical centre of the circles $x^2 + y^2 - 16x + 60 = 0$,$x^2 + y^2 - 12x + 27 = 0$,and $x^2 + y^2 - 12y + 8 = 0$ is

If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals