What is x^2-8x+y^2-4y+10=0 ?

The solution to x^2-8x+y^2-4y+10=0 is Circle with (a,b)=(4,2),r=sqrt(10)

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x^2-8x+y^2-4y+10=0

x^{2} - 8 x + y^{2} - 4 y + 10 = 0

(a, b) = (4, 2), r = 10

Solution steps

x^{2} - 8 x + y^{2} - 4 y + 10 = 0

Rewrite

x^{2} - 8 x + y^{2} - 4 y + 10 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 2)^{2} = (10)^{2}

Therefore the circle properties are:

(a, b) = (4, 2), r = 10

x^{2} + x + y^{2} = 16

x^{2} + y^{2} - \frac{608}{79} x - \frac{1386}{79} y - \frac{5640}{79} = 0

x^{(2)} + y^{(2)} + 4 x - 6 y + 12 = 0

(x + 1)^{2} + (y - 2)^{2} = 3^{2}

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

What is x^2-8x+y^2-4y+10=0 ?
The solution to x^2-8x+y^2-4y+10=0 is Circle with (a,b)=(4,2),r=sqrt(10)