The points of intersection of the circles $x^2 + y^2 = 25$ and $x^2 + y^2 - 8x + 7 = 0$ are

Consider the three circles: $S_{1} \equiv x^{2}+y^{2}-6x-6y+4=0$,$S_{2} \equiv x^{2}+y^{2}-2x-4y+3=0$,and $S_{3} \equiv x^{2}+y^{2}+2kx+2y+1=0$. If the radical centre of these three circles exists,then which of the following cannot be the value of $k$?

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 radical centre of three circles described on the three sides of a triangle as diameters is

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

If the circle $x^2+y^2+4x-6y+c=0$ bisects the circumference of the circle $x^2+y^2-6x+4y-12=0$,then $c$ is equal to