What is 4x^2+4y^2-20x+40y+5=0 ?

The solution to 4x^2+4y^2-20x+40y+5=0 is Circle with (a,b)=(5/2 ,-5),r=sqrt(30)

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4x^2+4y^2-20x+40y+5=0

4 x^{2} + 4 y^{2} - 20 x + 40 y + 5 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 20 x + 40 y + 5 = 0

Rewrite

4 x^{2} + 4 y^{2} - 20 x + 40 y + 5 = 0

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

x^{2} + y^{2} - 6 x - 8 y + 16 = 0

x^{2} + y^{2} + 10 x - 18 y + 70 = 0

x^{2} + y^{2} - 2 x - 80 = 0

5 x^{2} + 5 y^{2} - 25 = 0

x^{2} + y^{2} - 6 x + 2 y - 1 = 0

What is 4x^2+4y^2-20x+40y+5=0 ?
The solution to 4x^2+4y^2-20x+40y+5=0 is Circle with (a,b)=(5/2 ,-5),r=sqrt(30)