What is the center (x+1/3)^2+(y-5)^2=7 ?

The center (x+1/3)^2+(y-5)^2=7 is (-1/3 ,5)

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center (x+1/3)^2+(y-5)^2=7

center

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

(- \frac{1}{3}, 5)

Solution steps

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

Rewrite

(x + \frac{1}{3})^{2} + (y - 5)^{2} = 7

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - 5)^{2} = (7)^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, 5), r = 7

And the center is:

(- \frac{1}{3}, 5)

x^{2} - 4 x + y^{2} - 8 y = 20

(x - 3)^{2} + (y + 2)^{2} = \frac{16}{9}

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

3 x^{2} - 15 x = - 3 y^{2}

(x - 3)^{2} + y^{2} = 16

What is the center (x+1/3)^2+(y-5)^2=7 ?
The center (x+1/3)^2+(y-5)^2=7 is (-1/3 ,5)