What is 4x^2+4y^2-40x-8y+40=0 ?

The solution to 4x^2+4y^2-40x-8y+40=0 is Circle with (a,b)=(5,1),r=4

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4x^2+4y^2-40x-8y+40=0

4 x^{2} + 4 y^{2} - 40 x - 8 y + 40 = 0

(a, b) = (5, 1), r = 4

Solution steps

4 x^{2} + 4 y^{2} - 40 x - 8 y + 40 = 0

Rewrite

4 x^{2} + 4 y^{2} - 40 x - 8 y + 40 = 0

in the form of the standard circle equation

(x - 5)^{2} + (y - 1)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (5, 1), r = 4

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

(X - 1)^{2} + (Y + 1)^{2} = 1

(x - 2)^{2} + (y + 4)^{2} = 41

y^{2} + x^{2} - 8 x + 18 y + 72 = 0

x^{2} + y^{2} + 8 x + 4 y + 19 = 0

What is 4x^2+4y^2-40x-8y+40=0 ?
The solution to 4x^2+4y^2-40x-8y+40=0 is Circle with (a,b)=(5,1),r=4