What is 4x^2+4y^2-4x+8y+1=0 ?

The solution to 4x^2+4y^2-4x+8y+1=0 is Circle with (a,b)=(1/2 ,-1),r=1

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4x^2+4y^2-4x+8y+1=0

4 x^{2} + 4 y^{2} - 4 x + 8 y + 1 = 0

(a, b) = (\frac{1}{2}, - 1), r = 1

Solution steps

4 x^{2} + 4 y^{2} - 4 x + 8 y + 1 = 0

Rewrite

4 x^{2} + 4 y^{2} - 4 x + 8 y + 1 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- 1))^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 1), r = 1

(x + 3)^{2} + (y - \frac{4}{3})^{2} = 36

(x - 3)^{2} + (y - 5)^{2} = 4

(x - 1)^{2} + (y - 4)^{2} = 9

\frac{x ^{2}}{5} + \frac{y ^{2}}{5} = 53

x^{2} + y^{2} + 4 x + 8 y + 11 = 0

What is 4x^2+4y^2-4x+8y+1=0 ?
The solution to 4x^2+4y^2-4x+8y+1=0 is Circle with (a,b)=(1/2 ,-1),r=1