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

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

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

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

(a, b) = (- \frac{1}{2}, - 3), r = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, - 3), r = \frac{1}{2}

x^{2} + 4 x + y^{2} + 4 y - 8 = 0

(x - 8)^{2} + (y + 3)^{2} = 1

x^{2} + y^{2} - 4 x - 12 y - 6 = 0

(x + 3)^{2} + (y - 5)^{2} = 3

(x + 3)^{2} + (y - 5)^{2} = 9

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