What is 0.2x^2+x+0.2y^2-4y=1 ?

The solution to 0.2x^2+x+0.2y^2-4y=1 is Circle with (a,b)=(-5/2 ,10),r=(sqrt(445))/2

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0.2x^2+x+0.2y^2-4y=1

0.2 x^{2} + x + 0.2 y^{2} - 4 y = 1

(a, b) = (- \frac{5}{2}, 10), r = \frac{445}{2}

Solution steps

0.2 x^{2} + x + 0.2 y^{2} - 4 y = 1

Rewrite

0.2 x^{2} + x + 0.2 y^{2} - 4 y = 1

in the form of the standard circle equation

(x - (- \frac{5}{2}))^{2} + (y - 10)^{2} = (\frac{445}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 10), r = \frac{445}{2}

x^{2} + y^{2} - 5 x = 20 - 2 y

(x^{2} + 14 x 9) + (y^{2} + 12 y + 36) = 36 + 49 + 36

x^{2} + y^{2} - 10 x + 7 y + 30 = 0

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

(x - 3)^{2} + (y + 2)^{2} = 33

What is 0.2x^2+x+0.2y^2-4y=1 ?
The solution to 0.2x^2+x+0.2y^2-4y=1 is Circle with (a,b)=(-5/2 ,10),r=(sqrt(445))/2