What is (x+2)^2+(y-3/7)^2= 16/9 ?

The solution to (x+2)^2+(y-3/7)^2= 16/9 is Circle with (a,b)=(-2, 3/7),r= 4/3

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(x+2)^2+(y-3/7)^2= 16/9

(x + 2)^{2} + (y - \frac{3}{7})^{2} = \frac{16}{9}

(a, b) = (- 2, \frac{3}{7}), r = \frac{4}{3}

Solution steps

(x + 2)^{2} + (y - \frac{3}{7})^{2} = \frac{16}{9}

Rewrite

(x + 2)^{2} + (y - \frac{3}{7})^{2} = \frac{16}{9}

in the form of the standard circle equation

(x - (- 2))^{2} + (y - \frac{3}{7})^{2} = (\frac{4}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 2, \frac{3}{7}), r = \frac{4}{3}

x^{2} + y^{2} + 10 y + \frac{99}{4} = 0

(X + 3)^{2} + (Y - 0)^{2} = 9^{2}

(5 - x)^{2} + z^{2} = 4

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

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

What is (x+2)^2+(y-3/7)^2= 16/9 ?
The solution to (x+2)^2+(y-3/7)^2= 16/9 is Circle with (a,b)=(-2, 3/7),r= 4/3