Let the radius of both circles are ' $r$ '. Now, equation of circle with centre at $(2,3)$ is $S_{1} \equiv(x-2)^{2}+(y-3)^{2}=r^{2}$ and equation of circle with centre at $(5,6)$ is $S_{2} \equiv(x-5)^{2}+(y-6)^{2}=r^{2}$ Now, the equation common chord $\equiv$ Radical axis of $S_{1}$ and $S_{2}=0$ $\equiv\left(S_{1}-S_{2}\right)=0$ $\equiv\left[(x-2)^{2}\right]+\left[(y-3)^{2}\right]$ $\equiv x^{2}+y^{2}+4-4 x+9-6 x$ $-x^{2}-y^{2}-25-36+10 x+12 y=0$ $\equiv 6 x+6 y-48=0$ Common chord $\equiv x+y-8=0$