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

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

4 x^{2} + 4 y^{2} - 24 x - 12 y + 29 = 0

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

y^{2} + z^{2} = 25

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