Question

4X^2+4y^2-8X-20y+13=0

Answer 1

4x^2+4y^2-8x-20y+13=0

rearrange things a bit and you get

4x^2-8x + 4y^2-20y = -13
4(x^2-2x) + 4(y^2-5y) = -13

Now complete the squares:

4(x^2-2x+1) + 4(y^2-5y+25/4) = -13 + 4(1) + 4(25/4)
4(x-1)^2 + 4(y - 5/2)^2 = 16
(x-1)^2 + (y - 5/2)^2 = 4

Now does it look familiar?