What is 4x^2+4y^2-8x+16y+11=0 ?

The solution to 4x^2+4y^2-8x+16y+11=0 is Circle with (a,b)=(1,-2),r= 3/2

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4x^2+4y^2-8x+16y+11=0

4 x^{2} + 4 y^{2} - 8 x + 16 y + 11 = 0

(a, b) = (1, - 2), r = \frac{3}{2}

Solution steps

4 x^{2} + 4 y^{2} - 8 x + 16 y + 11 = 0

Rewrite

4 x^{2} + 4 y^{2} - 8 x + 16 y + 11 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 2))^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, - 2), r = \frac{3}{2}

x^{2} + y^{2} + 2 x - 2 y = 18

(x + 7)^{2} + (y + 4)^{2} = 49

x^{2} - 8 x + y^{2} - 9 = 0

4 x^{2} + 10 x + 4 y^{2} - y - 4 = 0

3 x^{2} + 3 y^{2} + 8 x + 6 y + 7 = 0

What is 4x^2+4y^2-8x+16y+11=0 ?
The solution to 4x^2+4y^2-8x+16y+11=0 is Circle with (a,b)=(1,-2),r= 3/2