Answer: \(B=(5,-5)\), and the tangent is \(y+1=\frac12(x-3)\).

The centre of the circle is \((4,-3)\), and its radius is \(\sqrt5\).

Substitute \(A=(3,-1)\) into the circle equation:

\((3-4)^2+(-1+3)^2=1+4=5.\)

So \(A\) lies on the circumference.

If \(AB\) is a diameter, then the centre is the midpoint of \(AB\). Let \(B=(x,y)\). Then

\(\frac{3+x}{2}=4,\quad \frac{-1+y}{2}=-3.\)

Hence

\(x=5,\quad y=-5.\)

So

\(B=(5,-5).\)

The gradient of the radius from the centre \((4,-3)\) to \(A=(3,-1)\) is

\(\frac{-1-(-3)}{3-4}=\frac{2}{-1}=-2.\)

The tangent is perpendicular to this radius, so its gradient is \(\frac12\). Through \(A=(3,-1)\), the tangent is

\(y+1=\frac12(x-3).\)