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

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

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

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

(a, b) = (4, 2), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (4, 2), r = 4

4 x^{2} + 4 y^{2} - 16 x + 20 y + 25 = 0

x^{2} + y^{2} + 8 x - 6 y = - 18

x^{2} + y^{2} - 32 x - 60 y + 1122 = 0

x^{2} + y^{2} = 1 5^{2}

(x + 2)^{2} + (y - 5)^{2} = 25

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