What is (x-1/4)^2+y^2= 1/16 ?

The solution to (x-1/4)^2+y^2= 1/16 is Circle with (a,b)=(1/4 ,0),r= 1/4

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(x-1/4)^2+y^2= 1/16

(x - \frac{1}{4})^{2} + y^{2} = \frac{1}{16}

(a, b) = (\frac{1}{4}, 0), r = \frac{1}{4}

Solution steps

(x - \frac{1}{4})^{2} + y^{2} = \frac{1}{16}

Rewrite

(x - \frac{1}{4})^{2} + y^{2} = \frac{1}{16}

in the form of the standard circle equation

(x - \frac{1}{4})^{2} + (y - 0)^{2} = (\frac{1}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{4}, 0), r = \frac{1}{4}

x^{2} + y^{2} + 10 x - 8 y - 8 = 0

- 1 = x^{2} + y^{2} - 6 x + 5

14 y + y^{2} + x^{2} = 12 x - 18

(x + 2)^{2} + (y - 3)^{2} = 40

(x + 2)^{2} + (y - 3)^{2} = 20

What is (x-1/4)^2+y^2= 1/16 ?
The solution to (x-1/4)^2+y^2= 1/16 is Circle with (a,b)=(1/4 ,0),r= 1/4