What is (x-(-5))^2+(y-(-6))^2= 2/3 ?

The solution to (x-(-5))^2+(y-(-6))^2= 2/3 is Circle with (a,b)=(-5,-6),r=sqrt(2/3)

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(x-(-5))^2+(y-(-6))^2= 2/3

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

(a, b) = (- 5, - 6), r = \frac{2}{3}

Solution steps

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

Rewrite

(x - (- 5))^{2} + (y - (- 6))^{2} = \frac{2}{3}

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 6))^{2} = (\frac{2}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 6), r = \frac{2}{3}

x^{2} + y^{2} - 6 x - 18 y + 41 = 0

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

x^{2} + (y + \frac{3}{5})^{2} = 27

x^{2} + y^{2} - 8 x + 12 y + 45 = 0

x^{2} + y^{2} - 8 x + 8 y + 16 = 0

What is (x-(-5))^2+(y-(-6))^2= 2/3 ?
The solution to (x-(-5))^2+(y-(-6))^2= 2/3 is Circle with (a,b)=(-5,-6),r=sqrt(2/3)