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

x^{2} + y^{2} - 8 x + 8 y - 49 = 0

x^{2} + y^{2} + 4 x = 96

x^{2} + y^{2} + 4 x + 8 y - 80 = 0

x^{2} - 2 x + y^{2} = \frac{416}{25}

(x + \frac{2}{5})^{2} + (y - \frac{3}{4})^{2} = 3

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)