What is the center x^2+y^2+10x-6y+15=0 ?

The center x^2+y^2+10x-6y+15=0 is (-5,3)

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center x^2+y^2+10x-6y+15=0

center

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

(- 5, 3)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 3)^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (- 5, 3), r = 19

And the center is:

(- 5, 3)

a^{2} + b^{2} = 36

(x + 5)^{2} + (y - 3)^{2} = 25

9 x^{2} + 9 y^{2} + 42 x - 54 y + 49 = 0

(x - (- 5))^{2} + (y - (7))^{2} = (11)^{2}

(x + 4)^{2} + (y + 1)^{2} = 9

What is the center x^2+y^2+10x-6y+15=0 ?
The center x^2+y^2+10x-6y+15=0 is (-5,3)