What is the answer and how do you do this problem?

We want to recast the equation into the standard form equation for a circle centered at (h, k) with radius r. That equation is

... (x -h)² + (y -k)² = r²

Start by completing the square for both x-terms and y-terms.

... x² - 4x + y² + 4y = k

To do that, add the squares of half the coefficients of the x- and y-terms.

... x² - 4x + (-2)² + y² + 4y + 2² = k + (-2)² + 2²

... (x -2)² + (y +2)² = k + 8 = r² . . . . . this is now equal to the square of the radius, so we have

... k + 8 = 6² = 36

Subtracting 8 gives

... k = 28 . . . . . . . matches selection D)

3 is greater then both of those numbers