What is the center x^2+y^2+6x-16y+48=0 ?

The center x^2+y^2+6x-16y+48=0 is (-3,8)

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center x^2+y^2+6x-16y+48=0

center

x^{2} + y^{2} + 6 x - 16 y + 48 = 0

(- 3, 8)

Solution steps

x^{2} + y^{2} + 6 x - 16 y + 48 = 0

Rewrite

x^{2} + y^{2} + 6 x - 16 y + 48 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 3, 8)

x^{2} + (y + 5)^{2} = 5

2 x^{2} + 2 y^{2} + 16 x - 32 y + 134 = 0

(x - 3)^{2} + (y + 4)^{2} = 2

(x - 0)^{2} + (y - 0)^{2} = 4

x^{2} + y^{2} + 10 x - 16 y + 75 = 0

What is the center x^2+y^2+6x-16y+48=0 ?
The center x^2+y^2+6x-16y+48=0 is (-3,8)