What is (x-1/2)^2+(y+1/4)^2=12 ?

The solution to (x-1/2)^2+(y+1/4)^2=12 is Circle with (a,b)=(1/2 ,-1/4),r=sqrt(12)

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(x-1/2)^2+(y+1/4)^2=12

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = 12

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = 12

Solution steps

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = 12

Rewrite

(x - \frac{1}{2})^{2} + (y + \frac{1}{4})^{2} = 12

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{1}{4}))^{2} = (12)^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{1}{4}), r = 12

x^{(2)} + y^{(2)} + 4 x - 10 y = - 16

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

4 x^{2} + 4 x + 4 y^{2} - 24 y - 27 = 0

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

x^{2} + y^{2} + 10 x - 56 = 0

What is (x-1/2)^2+(y+1/4)^2=12 ?
The solution to (x-1/2)^2+(y+1/4)^2=12 is Circle with (a,b)=(1/2 ,-1/4),r=sqrt(12)