What is (x+1)^2+y^2= 4/9 ?

The solution to (x+1)^2+y^2= 4/9 is Circle with (a,b)=(-1,0),r= 2/3

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(x+1)^2+y^2= 4/9

(x + 1)^{2} + y^{2} = \frac{4}{9}

(a, b) = (- 1, 0), r = \frac{2}{3}

Solution steps

(x + 1)^{2} + y^{2} = \frac{4}{9}

Rewrite

(x + 1)^{2} + y^{2} = \frac{4}{9}

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 0)^{2} = (\frac{2}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, 0), r = \frac{2}{3}

(x - 3)^{2} + (y - 0)^{2} = 2^{2}

x^{2} + y^{2} + 10 y + 19 = 0

x^{2} + y^{2} + 4 x - 4 y - 17 = 0

x^{2} + y^{2} - 4 x + 2 y - 15 = 0

x^{2} + y^{2} - 10 x - 4 y - 56 = 0

What is (x+1)^2+y^2= 4/9 ?
The solution to (x+1)^2+y^2= 4/9 is Circle with (a,b)=(-1,0),r= 2/3