What is (x-1/(2(3/2)))^2+y^2=(1/(2(3/2)))^2 ?

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

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(x-1/(2(3/2)))^2+y^2=(1/(2(3/2)))^2

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 25 x = - 1 - 3 y

(x - 11)^{2} + (y - 8)^{2} = 268

4 x^{2} - 24 x + 4 y^{2} + 20 y = 39

x^{2} + (y + 1)^{2} = 64

\frac{1}{2} x^{2} + \frac{1}{2} y^{2} = 1

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