One approach is to use ezplot:
ezplot('x^2+y^2-4*x-8*y+10-2*log(2)')
By default, this plots over the domain -2π < x < 2π, -2π < y < 2π. But you'll see that needs adjusting. Add axis limits:
ezplot('x^2+y^2-4*x-8*y+10-2*log(2)',[-2 6 0 8])
Finally, type
axis equal
and you'll see that it's actually a square.
An alternative is that you can complete the square (see, e.g., Completing the Square: Ellipse Equations), getting (x-2)^2+(y-4)^2=10+2*log(2). This is a circle with center at (2,4) and radius sqrt(10+2*log(2)) (about 3.4).
