Step 1: Find the coordinates of D.

The midpoint D of AB is given by:

\(D = \left( \frac{1 + 9}{2}, \frac{3 + (-1)}{2} \right) = (5, 1)\)

Step 2: Write an equation for CD² = 20.

The distance squared from C to D is:

\((x - 5)^2 + (y - 1)^2 = 20\)

Step 3: Find an equation for AC = BC and simplify.

The distances squared are:

\((x - 1)^2 + (y - 3)^2 = (x - 9)^2 + (y + 1)^2\)

Expanding both sides:

\(x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 - 18x + 81 + y^2 + 2y + 1\)

Simplifying gives:

\(-2x - 6y + 10 = -18x + 2y + 82\)

\(16x - 8y = 72\)

\(y = 2x - 9\)

Step 4: Find the possible coordinates of C.

Substitute \(y = 2x - 9\) into \((x - 5)^2 + (y - 1)^2 = 20\):

\((x - 5)^2 + (2x - 10)^2 = 20\)

\((x - 5)^2 + (2x - 10)^2 = 20\)

\(5x^2 - 50x + 105 = 0\)

Solving gives \(x = 3\) or \(x = 7\).

For \(x = 3\), \(y = 2(3) - 9 = -3\).

For \(x = 7\), \(y = 2(7) - 9 = 5\).

Thus, the possible coordinates of C are (3, -3) and (7, 5).