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

The solution to 4x^2+4y^2-x+2y=11 is Circle with (a,b)=(1/8 ,-1/4),r=(sqrt(181))/8

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4x^2+4y^2-x+2y=11

4 x^{2} + 4 y^{2} - x + 2 y = 11

(a, b) = (\frac{1}{8}, - \frac{1}{4}), r = \frac{181}{8}

Solution steps

4 x^{2} + 4 y^{2} - x + 2 y = 11

Rewrite

4 x^{2} + 4 y^{2} - x + 2 y = 11

in the form of the standard circle equation

(x - \frac{1}{8})^{2} + (y - (- \frac{1}{4}))^{2} = (\frac{181}{8})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{8}, - \frac{1}{4}), r = \frac{181}{8}

x^{2} + y^{2} - 4 x + 8 y - 12 = 0

x^{2} + y^{2} + 7.2 x + 8.6 y = 2.19

x^{2} + (y - 3)^{2} = 20

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

(x + 20)^{2} + (y - 30)^{2} = 225

What is 4x^2+4y^2-x+2y=11 ?
The solution to 4x^2+4y^2-x+2y=11 is Circle with (a,b)=(1/8 ,-1/4),r=(sqrt(181))/8