What is -4y^2-4x^2-20x+32y+81=0 ?

The solution to -4y^2-4x^2-20x+32y+81=0 is Circle with (a,b)=(-5/2 ,4),r=sqrt(85/2)

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-4y^2-4x^2-20x+32y+81=0

- 4 y^{2} - 4 x^{2} - 20 x + 32 y + 81 = 0

(a, b) = (- \frac{5}{2}, 4), r = \frac{85}{2}

Solution steps

- 4 y^{2} - 4 x^{2} - 20 x + 32 y + 81 = 0

Rewrite

- 4 y^{2} - 4 x^{2} - 20 x + 32 y + 81 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - 4)^{2} = (\frac{85}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 4), r = \frac{85}{2}

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

x^{2} + y^{2} + 2 x - 14 y = - 25

x^{2} + y^{2} + 4 x - 6 y - 16 = 0

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

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

What is -4y^2-4x^2-20x+32y+81=0 ?
The solution to -4y^2-4x^2-20x+32y+81=0 is Circle with (a,b)=(-5/2 ,4),r=sqrt(85/2)