What is 4x^2+4y^2-20x-32y+81=0 ?

The solution to 4x^2+4y^2-20x-32y+81=0 is Circle with (a,b)=(5/2 ,4),r=sqrt(2)

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4x^2+4y^2-20x-32y+81=0

4 x^{2} + 4 y^{2} - 20 x - 32 y + 81 = 0

(a, b) = (\frac{5}{2}, 4), r = 2

Solution steps

4 x^{2} + 4 y^{2} - 20 x - 32 y + 81 = 0

Rewrite

4 x^{2} + 4 y^{2} - 20 x - 32 y + 81 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, 4), r = 2

x^{2} + y^{2} + 6 x + 4 y - 3 = 0

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

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

3 x^{2} + 3 y^{2} - 10 = 0

x^{2} + y^{2} - 3 x + 5 y - 14 = 0

What is 4x^2+4y^2-20x-32y+81=0 ?
The solution to 4x^2+4y^2-20x-32y+81=0 is Circle with (a,b)=(5/2 ,4),r=sqrt(2)