What is x^2+y^2-4x+14y+40=0 ?

The solution to x^2+y^2-4x+14y+40=0 is Circle with (a,b)=(2,-7),r=sqrt(13)

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x^2+y^2-4x+14y+40=0

x^{2} + y^{2} - 4 x + 14 y + 40 = 0

(a, b) = (2, - 7), r = 13

Solution steps

x^{2} + y^{2} - 4 x + 14 y + 40 = 0

Rewrite

x^{2} + y^{2} - 4 x + 14 y + 40 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 7))^{2} = (13)^{2}

Therefore the circle properties are:

(a, b) = (2, - 7), r = 13

x^{2} + y^{2} + 2 x - 6 y + 7 = 0

x^{2} + y^{2} + x + y - \frac{1}{2} = 0

x^{2} + y^{2} + 6 x + 4 y = 0

- 9 = - y^{2} - x^{2}

(x - 11)^{2} + (y - 14)^{2} = 8

What is x^2+y^2-4x+14y+40=0 ?
The solution to x^2+y^2-4x+14y+40=0 is Circle with (a,b)=(2,-7),r=sqrt(13)