What is the center X^2+Y^2+40X-10Y+55=0 ?

The center X^2+Y^2+40X-10Y+55=0 is (-20,5)

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center X^2+Y^2+40X-10Y+55=0

center

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

(- 20, 5)

Solution steps

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

Rewrite

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

in the form of the standard circle equation

(X - (- 20))^{2} + (Y - 5)^{2} = (370)^{2}

Therefore the circle properties are:

(a, b) = (- 20, 5), r = 370

And the center is:

(- 20, 5)

x^{2} + y^{2} + 6 x - 4 y - 5 = 0

(x + 2)^{2} + (y + 3)^{2} = 81

3 x^{2} + 3 y^{2} - 18 x + 6 y + 2 = 0

(x - 13)^{2} + (y + 12)^{2} = 64

x^{2} + y^{2} - \frac{2}{3} x - \frac{5}{3} y - \frac{5}{9} = 0

What is the center X^2+Y^2+40X-10Y+55=0 ?
The center X^2+Y^2+40X-10Y+55=0 is (-20,5)