Explanation

(c)
Let the equation of circle be
$x^{2} + y^{2} + 2hx + 2ky + c = 0$ …(i)
The locus of whose centre is to be obtained, since the circle cuts
$x^{2} + y^{2} + 4x - 6y + 9 = 0$ …(ii)
And $x^{2} + y^{2} - 4x + 6y + 4 = 0$ …(iii)
Orthogonally, then
$2h(2) + 2k( - 3) = c + 9$
$\Rightarrow 4h - 6k = c + 9$
And $2h( - 2) + 2k(3) = c + 4$
$\Rightarrow \ - 4h + 6k = c + 4$
On solving Eqs. (iv) and (v), we get
$c + 9 = - c - 4$
$\Rightarrow 2c = - 13$ …(vi)
On putting the value of $c$ in Eq. (iv)
$\Rightarrow 8h - 12k = 5$ …(vii)
Centre of the given circle is $( - h,\ - k)$
$\therefore$ Locus of $( - h,\ - k)$ from Eq. (vii) is
$8( - x) - 12( - y) = 5$
$\Rightarrow 8x - 12y + 5 = 0$