What is the center x^2+y^2-2x+8y-47=0 ?

The center x^2+y^2-2x+8y-47=0 is (1,-4)

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Find the Center and Radius x^2+y^2-2x+8y-47=0

center

x^{2} + y^{2} - 2 x + 8 y - 47 = 0

(1, - 4)

Solution steps

x^{2} + y^{2} - 2 x + 8 y - 47 = 0

Rewrite

x^{2} + y^{2} - 2 x + 8 y - 47 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 4))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (1, - 4), r = 8

And the center is:

(1, - 4)

\frac{( x ^{2} )}{73} - \frac{( y ^{2} )}{19} = 1

p (x) = 21 + 24 x + 6 x^{2}

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

h (x) = x^{2} - 14 x + 6

8 x^{2} - 48 x + 65