What is the center x^2+y^2+2x-4y-15=0 ?

The center x^2+y^2+2x-4y-15=0 is (-1,2)

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center x^2+y^2+2x-4y-15=0

center

x^{2} + y^{2} + 2 x - 4 y - 15 = 0

(- 1, 2)

Solution steps

x^{2} + y^{2} + 2 x - 4 y - 15 = 0

Rewrite

x^{2} + y^{2} + 2 x - 4 y - 15 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 2)^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (- 1, 2), r = 25

And the center is:

(- 1, 2)

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

(x + 2)^{2} + y^{2} = 9

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0

2 x^{2} + 2 y^{2} + 3 x + 5 y - 4 = 0

(x - 1)^{2} + (y - 2)^{2} = 64