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 $\lambda=$

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

If the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,is $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$,then $\lambda=$

If the cosine of the angle between the two circles $x^2+y^2+2x+4y-3=0$ and $x^2+y^2+2kx-2y-1=0$ is $\frac{1}{2\sqrt{3}}$,then $k^2=$

If the circle $S=0$ intersects the three circles $S_1 \equiv x^2+y^2+4x-7=0$,$S_2 \equiv x^2+y^2+y=0$ and $S_3 \equiv x^2+y^2+\frac{3}{2}x+\frac{5}{2}y-\frac{9}{2}=0$ orthogonally,then the radical axis of $S=0$ and $S_1=0$ is

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is: