If the circles $x^2+y^2-2 \lambda x-2 y-7=0$ and $3(x^2+y^2)-8 x+29 y=0$ are orthogonal,then $\lambda=$

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

Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. If the radius of the circle $S=0$ is $1$,then find the value of $g+f$.

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2x+6y=0$,$S_2: x^2+y^2+2gx-2y+6=0$,and $S_3: x^2+y^2-12x+2fy+3=0$. If $S_2$ and $S_3$ intersect orthogonally,then $(g, f) =$

$A$ circle $S$ given by $x^2+y^2-14x+6y+33=0$ cuts the $X$-axis at $A$ and $B$ $(OB > OA)$. $C$ is the midpoint of $AB$. $L$ is a line through $C$ with slope $-1$. If $L$ is the diameter of a circle $S^{\prime}$ and also the radical axis of the circles $S$ and $S^{\prime}$,then the equation of the circle $S^{\prime}$ 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: