What is 4x^2+4y^2+20x-8y+9=0 ?

The solution to 4x^2+4y^2+20x-8y+9=0 is Circle with (a,b)=(-5/2 ,1),r=sqrt(5)

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4x^2+4y^2+20x-8y+9=0

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(a, b) = (- \frac{5}{2}, 1), r = 5

Solution steps

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

Rewrite

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - 1)^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 1), r = 5

\frac{1}{2} y^{2} = - \frac{1}{2} x^{2} + 12.5

(x - 2)^{2} + (y + 3)^{2} = 11

x^{2} + y^{2} + 16 x - 14 y + 49 = 0

x^{2} + y^{2} + 2 x + 8 y - 19 = 0

5 x^{2} + 5 y^{2} - 2 x - 30 y + 42 = 0

What is 4x^2+4y^2+20x-8y+9=0 ?
The solution to 4x^2+4y^2+20x-8y+9=0 is Circle with (a,b)=(-5/2 ,1),r=sqrt(5)