First I found the equation of the line normal to $\displaystyle 3x-y-1=0$ and passing through A. The centre of the circle lies on this line. Then I found the equation of the perpendicular bisector of the chord AB. The centre of the circle also lies on this line. Solving for the point of intersection, I determined the centre of the circle to be $\displaystyle (4,1).$ Finally, I found the radius of the circle, which is $\displaystyle \sqrt{10}.$ Hence the circle is $\displaystyle (x-4)^2+(y-1)^2=10.$