What is the center 30=x^2-10x+y^2+6y ?

The center 30=x^2-10x+y^2+6y is (5,-3)

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center 30=x^2-10x+y^2+6y

center

30 = x^{2} - 10 x + y^{2} + 6 y

(5, - 3)

Solution steps

30 = x^{2} - 10 x + y^{2} + 6 y

Rewrite

30 = x^{2} - 10 x + y^{2} + 6 y

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 3))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (5, - 3), r = 8

And the center is:

(5, - 3)

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

x^{2} + y^{2} + 10 x - 18 y + 70 = 0

8 x^{2} + 8 y^{2} = 120

4 x^{2} - 8 x + 4 y^{2} + 12 y - 243 = 0

2 x^{2} + 2 y^{2} + 20 x + 16 y + 9 = 0