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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x + 3)^{2} + (y - 2)^{2} = \frac{25}{4}

x^{2} + y^{2} = 13

x^{2} + y^{2} + 6 x + 5 y = \frac{45}{4}

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

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

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