What is x^2+y^2-4x+11-84=0 ?

The solution to x^2+y^2-4x+11-84=0 is Circle with (a,b)=(2,0),r=sqrt(77)

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x^2+y^2-4x+11-84=0

x^{2} + y^{2} - 4 x + 11 - 84 = 0

(a, b) = (2, 0), r = 77

Solution steps

x^{2} + y^{2} - 4 x + 11 - 84 = 0

Rewrite

x^{2} + y^{2} - 4 x + 11 - 84 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - 0)^{2} = (77)^{2}

Therefore the circle properties are:

(a, b) = (2, 0), r = 77

2 x^{2} + 2 y^{2} - 12 x + 4 y + 18 = 0

(x + 7)^{2} + y^{2} = 16

(x + 1)^{2} + (y - 1)^{2} = 9 (x - 2)^{2} + 9 (y + 1)^{2}

(x - 9)^{2} + (y - 8)^{2} = 81

x^{2} + y^{2} - 16 x + 8 y + 55 = 0

What is x^2+y^2-4x+11-84=0 ?
The solution to x^2+y^2-4x+11-84=0 is Circle with (a,b)=(2,0),r=sqrt(77)