What is the center x^2+y^2-8y=0 ?

The center x^2+y^2-8y=0 is (0,4)

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center x^2+y^2-8y=0

center

x^{2} + y^{2} - 8 y = 0

(0, 4)

Solution steps

x^{2} + y^{2} - 8 y = 0

Rewrite

x^{2} + y^{2} - 8 y = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{4 ^{2}} + \frac{( y - 4 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 4), a = 4, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 4), b = 4, a = 4

And the center is:

(0, 4)

\frac{( x - 1 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{5} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

\frac{x ^{2}}{5} + \frac{y ^{2}}{9} = 1

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

8 x^{2} + 2 y^{2} + 16 x = 0