What is x^2+2x*3+3^2+y^2-2y*2/3+(2/3)^2=4^2 ?

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

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x^2+2x3+3^2+y^2-2y2/3+(2/3)^2=4^2

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} - 8 x + 16 + y^{2} - 18 y + 81 = 48

(x + 3)^{2} + (y - 2)^{2} = 49

x^{2} - 8 x + 16 + y^{2} - 18 y + 81 = 16

x^{2} + y^{2} - 6 x - 2 y + 9 = 0

- y^{2} = x^{2} - 49

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