What is the center (x-3)^2+(y+4)^2= 16/9 ?

The center (x-3)^2+(y+4)^2= 16/9 is (3,-4)

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center (x-3)^2+(y+4)^2= 16/9

center

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

(3, - 4)

Solution steps

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

Rewrite

(x - 3)^{2} + (y + 4)^{2} = \frac{16}{9}

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(3, - 4)

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

(x + 2)^{2} + (y - 3)^{2} = 15

(x + 5)^{2} + (y + 2)^{2} = 29

x^{2} + y^{2} - 16 x + 12 y + 64 = 0

x^{2} + 2 x - 4 y + y^{2} = 76

What is the center (x-3)^2+(y+4)^2= 16/9 ?
The center (x-3)^2+(y+4)^2= 16/9 is (3,-4)