What is x^2+y^2-16x+40y+67=0 ?

The solution to x^2+y^2-16x+40y+67=0 is Circle with (a,b)=(8,-20),r=sqrt(397)

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x^2+y^2-16x+40y+67=0

x^{2} + y^{2} - 16 x + 40 y + 67 = 0

(a, b) = (8, - 20), r = 397

Solution steps

x^{2} + y^{2} - 16 x + 40 y + 67 = 0

Rewrite

x^{2} + y^{2} - 16 x + 40 y + 67 = 0

in the form of the standard circle equation

(x - 8)^{2} + (y - (- 20))^{2} = (397)^{2}

Therefore the circle properties are:

(a, b) = (8, - 20), r = 397

x^{2} + y^{2} + 8 x + 2 y + 9 = 0

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

(x - 0)^{2} + (y - 0)^{2} = 7^{2}

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

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

What is x^2+y^2-16x+40y+67=0 ?
The solution to x^2+y^2-16x+40y+67=0 is Circle with (a,b)=(8,-20),r=sqrt(397)