Question 116023
x^2+y^2-8x+4y+20=0
--------
x^2-8x+16 + y^2+4y+4 = -20 + 16 +4
(x-4)^2 + (x+2)^2 = 0
-----------------------
This is not a circle as you said.
-----------------------
Cheers,
Stan H.