What is the center (x+3)^2+(y+7/3)^2= 49/9 ?

The center (x+3)^2+(y+7/3)^2= 49/9 is (-3,-7/3)

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center (x+3)^2+(y+7/3)^2= 49/9

center

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

(- 3, - \frac{7}{3})

Solution steps

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

Rewrite

(x + 3)^{2} + (y + \frac{7}{3})^{2} = \frac{49}{9}

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 3, - \frac{7}{3})

x^{2} + y^{2} + 14 x - 16 + 64 = 0

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

2 x^{2} + 2 y^{2} - 5 x + 4 y - 1 = 0

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

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

What is the center (x+3)^2+(y+7/3)^2= 49/9 ?
The center (x+3)^2+(y+7/3)^2= 49/9 is (-3,-7/3)