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

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

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

center

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

(3, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(3, - 2)

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

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

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

x^{2} - 4 x + y^{2} = 21

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

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