If a system of circles pass through (2, 3) and cuts the circles$x^2+ y^2 = 12$ orthogonally, then the equation of the locus of the centres of that system of circles, is

$$\begin{gathered} \mathrm{Let} \mathrm{S}={\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ \mathrm{be} \mathrm{a} \mathrm{circle} \mathrm{passing} \mathrm{through} (2, 3) then \\ 4\mathrm{g}+6\mathrm{f}+13+\mathrm{c}=0 \to \left(1\right) \\ It is o\mathrm{rthogonal} \mathrm{to} {\mathrm{x}}^2+{\mathrm{y}}^2=12 then \\ \Rightarrow 2gg\text{'}+2ff\text{'}= \hspace{0.17em}c+c^1 \\ \Rightarrow c+c^1=0 \\ \Rightarrow c-12=0 \\ \Rightarrow c=12 \\ from\left(1\right)\Rightarrow 4\mathrm{g}+6\mathrm{f}+13+12=0 \\ 4\mathrm{g}+6\mathrm{f}+25=0 \\ \Rightarrow 4(-g)+6(-f)-25=0 \\ locus of it\text{'}s centre is 4\mathrm{x}+6\mathrm{y}-25=0 \end{gathered}$$