Let the centre of the circle be $(h, k).$ Since the centre lies on the line $y = x - 1$ $\therefore k-h-1\,...\left(i\right)$ Since the circle passes through the point $(7, 3)$, therefore, the distance of the centre from this point is the radius of the circle. $\therefore 3=\sqrt{\left(h-7\right)^{2}+\left(k-3\right)^{2}}$ $\Rightarrow 3=\sqrt{\left(h-7\right)^{2}+\left(h-1-3\right)^{2}}\,\left[using \left(i\right)\right]$ $\Rightarrow h = 7$ or $h = 4$ For $h = 7$, we get $k = 6$ and for $h = 4$, we get $k = 3$ Hence, the circles which satisfy the given conditions are: $\left(x-7\right)^{2}+\left(y-6\right)^{2}=9$ or $x^{2}+y^{2}-14x+12y+76=0$ and $\left(x-4\right)^{2}+\left(y-3\right)^{2}=9$ or $x^{2}+y^{2}-8x-6y+16=0$