What is the center (x-2)^2+(y+3)^2= 16/9 ?

The center (x-2)^2+(y+3)^2= 16/9 is (2,-3)

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center (x-2)^2+(y+3)^2= 16/9

center

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

(2, - 3)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 3), r = \frac{4}{3}

And the center is:

(2, - 3)

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

x^{2} + (y - 2)^{2} = 36

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

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

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

What is the center (x-2)^2+(y+3)^2= 16/9 ?
The center (x-2)^2+(y+3)^2= 16/9 is (2,-3)