What is 8x^2+8y^2-x-3y-11=0 ?

The solution to 8x^2+8y^2-x-3y-11=0 is Circle with (a,b)=(1/16 , 3/16),r=(sqrt(181))/(8sqrt(2))

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8x^2+8y^2-x-3y-11=0

8 x^{2} + 8 y^{2} - x - 3 y - 11 = 0

(a, b) = (\frac{1}{16}, \frac{3}{16}), r = \frac{181}{8 2}

Solution steps

8 x^{2} + 8 y^{2} - x - 3 y - 11 = 0

Rewrite

8 x^{2} + 8 y^{2} - x - 3 y - 11 = 0

in the form of the standard circle equation

(x - \frac{1}{16})^{2} + (y - \frac{3}{16})^{2} = (\frac{181}{8 2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{16}, \frac{3}{16}), r = \frac{181}{8 2}

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

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

(x - 2)^{2} + y^{2} - 64 = 0

x^{2} + y^{2} + 8 x + 15 = 0

(x - 7)^{2} + (y + 7)^{2} = 64

What is 8x^2+8y^2-x-3y-11=0 ?
The solution to 8x^2+8y^2-x-3y-11=0 is Circle with (a,b)=(1/16 , 3/16),r=(sqrt(181))/(8sqrt(2))