Let
\[
S\equiv x^2+y^2-6x+4y+c=0,\qquad
S'\equiv x^2+y^2-4x+6y+9=0,
\]
\[
S''\equiv x^2+y^2+5x+3y+k=0
\]
be three circles. If the angles of intersection of the circle \(S'=0\) with the circles \(S=0\) and \(S''=0\) are respectively
\[
\frac{\pi}{4}
\quad\text{and}\quad
\frac{\pi}{2},
\]
then \(c+k=\)
Show Hint
For two intersecting circles,
\[
\boxed{
\cos\theta
=
\frac{r_1^2+r_2^2-d^2}{2r_1r_2}
}
\]
where \(d\) is the distance between their centres.
For orthogonal circles,
\[
\boxed{r_1^2+r_2^2=d^2.}
\]