What is x^2-14x+y^2+8y-40=0 ?

The solution to x^2-14x+y^2+8y-40=0 is Circle with (a,b)=(7,-4),r=sqrt(105)

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x^2-14x+y^2+8y-40=0

x^{2} - 14 x + y^{2} + 8 y - 40 = 0

(a, b) = (7, - 4), r = 105

Solution steps

x^{2} - 14 x + y^{2} + 8 y - 40 = 0

Rewrite

x^{2} - 14 x + y^{2} + 8 y - 40 = 0

in the form of the standard circle equation

(x - 7)^{2} + (y - (- 4))^{2} = (105)^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 105

x^{2} + y^{2} - 6 x - 10 y - 15 = 0

x^{2} + y^{2} + 6 x - 12 y + 20 = 0

x^{2} + y^{2} + 6 x - 8 y + 9 = 0

x^{2} + y^{2} - 5 x = 0

2 x^{2} + 2 y^{2} - 2 x - 6 y - 13 = 0

What is x^2-14x+y^2+8y-40=0 ?
The solution to x^2-14x+y^2+8y-40=0 is Circle with (a,b)=(7,-4),r=sqrt(105)