A circle has radius 3 and its centre lies on the line $y=x-1$. The equation of the circle, if it passes through $(7,3)$, is

Let the centre of the circle be $(h, k)$. Since the centre lies on the line $y=x-1$ $\therefore k=h-1 ......(i)$ Since the circle passes through the point $(7,3)$, therefore, the distance of the centre from this point is the radius of the circle. $\therefore \quad 3=\sqrt{(h-7)^{2}+(k-3)^{2}}$ $\Rightarrow \quad 3=\sqrt{(h-7)^{2}+(h-1-3)^{2}}$ $\Rightarrow \quad h=7$ 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: $(x-7)^{2}+(y-6)^{2}=9$ or $x^{2}+y^{2}-14 x+12 y+76=0$ and $(x-4)^{2}+(y-3)^{2}=9$ or $$ x^{2}+y^{2}-8 x-6 y+16=0 $$