What is (x-0)^2+(y-1/2)^2=(1/2)^2 ?

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

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(x-0)^2+(y-1/2)^2=(1/2)^2

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

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

(x + 3)^{2} + (y + 2)^{2} = 25

36 = \frac{1}{49} (x^{2} + y^{2})

4 x^{2} + 4 y^{2} - 24 y - 13 = 0

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