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

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

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

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

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = 1

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = 1

2 x^{2} + 2 y^{2} - 16 y + 24 = 0

x^{2} + y^{2} - x = 2

(x - 4)^{(2)} + (y + 10)^{(2)} = 144

x^{2} + 6 x + 9 + y^{2} - 2 y + 1 = 54 + 9 + 1

x^{2} = - y^{2} + 10 y + 5

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