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

The solution to x^2+(y+3/2)^2=4 is Circle with (a,b)=(0,-3/2),r=2

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

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

(a, b) = (0, - \frac{3}{2}), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{3}{2}))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{3}{2}), r = 2

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

x^{2} - 8 x + y^{2} + 6 y = - 9

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

9 x^{2} + 9 y^{2} + 72 x + 6 y + 1 = 0

x^{2} + (y - 3)^{2} = 16

What is x^2+(y+3/2)^2=4 ?
The solution to x^2+(y+3/2)^2=4 is Circle with (a,b)=(0,-3/2),r=2