What is x^2-4x+y^2+14y=11 ?

The solution to x^2-4x+y^2+14y=11 is Circle with (a,b)=(2,-7),r=8

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x^2-4x+y^2+14y=11

x^{2} - 4 x + y^{2} + 14 y = 11

(a, b) = (2, - 7), r = 8

Solution steps

x^{2} - 4 x + y^{2} + 14 y = 11

Rewrite

x^{2} - 4 x + y^{2} + 14 y = 11

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 7))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (2, - 7), r = 8

x^{2} + y^{2} - 14 x + 18 y - 12 = 0

(x - 1)^{2} + y^{2} \leq 1

(x - 2)^{2} + (y + 7)^{2} = 49

x^{2} + y^{2} + 8 x + 12 y = 0

(x + 3)^{2} + (y - 4)^{2} = 7

What is x^2-4x+y^2+14y=11 ?
The solution to x^2-4x+y^2+14y=11 is Circle with (a,b)=(2,-7),r=8