What is (x+1/2)^2+(y-3/4)^2=4 ?

The solution to (x+1/2)^2+(y-3/4)^2=4 is Circle with (a,b)=(-1/2 , 3/4),r=2

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(x+1/2)^2+(y-3/4)^2=4

(x + \frac{1}{2})^{2} + (y - \frac{3}{4})^{2} = 4

(a, b) = (- \frac{1}{2}, \frac{3}{4}), r = 2

Solution steps

(x + \frac{1}{2})^{2} + (y - \frac{3}{4})^{2} = 4

Rewrite

(x + \frac{1}{2})^{2} + (y - \frac{3}{4})^{2} = 4

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - \frac{3}{4})^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{3}{4}), r = 2

(x - 3)^{2} + (y - 2)^{2} = 13

x^{2} + y^{2} - 9 x + 4 y - 94 = 0

(x - \frac{2}{3})^{2} + y^{2} = 36

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

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

What is (x+1/2)^2+(y-3/4)^2=4 ?
The solution to (x+1/2)^2+(y-3/4)^2=4 is Circle with (a,b)=(-1/2 , 3/4),r=2