What is (x-2/3)^2+(y-4/3)^2= 20/9 ?

The solution to (x-2/3)^2+(y-4/3)^2= 20/9 is Circle with (a,b)=(2/3 , 4/3),r=sqrt(20/9)

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(x-2/3)^2+(y-4/3)^2= 20/9

(x - \frac{2}{3})^{2} + (y - \frac{4}{3})^{2} = \frac{20}{9}

(a, b) = (\frac{2}{3}, \frac{4}{3}), r = \frac{20}{9}

Solution steps

(x - \frac{2}{3})^{2} + (y - \frac{4}{3})^{2} = \frac{20}{9}

Rewrite

(x - \frac{2}{3})^{2} + (y - \frac{4}{3})^{2} = \frac{20}{9}

in the form of the standard circle equation

(x - \frac{2}{3})^{2} + (y - \frac{4}{3})^{2} = (\frac{20}{9})^{2}

Therefore the circle properties are:

(a, b) = (\frac{2}{3}, \frac{4}{3}), r = \frac{20}{9}

(x^{2} - 2 x) + (y^{2} - 2 y) = 7

x^{2} + y^{2} - 1 = x

3 x^{2} - 12 x + 3 y^{2} - 6 y - 12 = 0

x^{2} + y^{2} + 4 x + 10 y + 25 = 0

x^{2} + y^{2} + 2 x - 10 y - 4 = 0

What is (x-2/3)^2+(y-4/3)^2= 20/9 ?
The solution to (x-2/3)^2+(y-4/3)^2= 20/9 is Circle with (a,b)=(2/3 , 4/3),r=sqrt(20/9)