What is 4x^2+4y^2-20x+40y=-5 ?

The solution to 4x^2+4y^2-20x+40y=-5 is Circle with (a,b)=(5/2 ,-5),r=sqrt(30)

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4x^2+4y^2-20x+40y=-5

4 x^{2} + 4 y^{2} - 20 x + 40 y = - 5

(a, b) = (\frac{5}{2}, - 5), r = 30

Solution steps

4 x^{2} + 4 y^{2} - 20 x + 40 y = - 5

Rewrite

4 x^{2} + 4 y^{2} - 20 x + 40 y = - 5

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - (- 5))^{2} = (30)^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, - 5), r = 30

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

(x - 1)^{2} + (y - 2)^{2} = 10

x^{2} + y^{2} - 14 x + 38 = 0

x^{2} + y^{2} + 4 x - 5 = 0

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

What is 4x^2+4y^2-20x+40y=-5 ?
The solution to 4x^2+4y^2-20x+40y=-5 is Circle with (a,b)=(5/2 ,-5),r=sqrt(30)