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 $y=x-1$ 

$\therefore k=h-1$

Since the circle passes through the point (7, 3), therefore, the distance of the centre from this point is the radius of the circle.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3=\sqrt{(h-7{)}^2+(k-3{)}^2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3=\sqrt{(h-7{)}^2+(h-1-3{)}^2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h=7\text{ or }h=4\end{array}$

For $h=7,\text{ we get }k=6$

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 $x^2+y^2-14x+12y+76=0$

and $(x-4{)}^2+(y-3{)}^2=9$

or $x^2+y^2-8x-6y+16=0$