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=5

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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 = 5

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} = 5^{2}

Therefore the circle properties are:

(a, b) = (7, - 4), r = 5

9 x^{2} + 9 y^{2} + 42 x + 84 y + 65 = 0

4 x^{2} + 4 y^{2} - 16 x - 24 y + 51 = 0

2 x^{2} + 2 y^{2} - 2 x + 6 y = 45

9 x^{2} + 9 y^{2} - 6 x + 24 y = 19

2 x^{2} + 2 y^{2} + 12 x - 2 y - 3 = 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=5