What is (x+pi)^2+(y-2pi)^2=8 ?

The solution to (x+pi)^2+(y-2pi)^2=8 is Circle with (a,b)=(-pi,2pi),r=sqrt(8)

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(x+pi)^2+(y-2pi)^2=8

(x + π)^{2} + (y - 2 π)^{2} = 8

(a, b) = (- π, 2 π), r = 8

Solution steps

(x + π)^{2} + (y - 2 π)^{2} = 8

Rewrite

(x + π)^{2} + (y - 2 π)^{2} = 8

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- π, 2 π), r = 8

18 y - 15 = - x^{2} - y^{2}

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

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

(x - 4)^{2} + (y + 3)^{2} = 5

x^{2} + z^{2} = 36

What is (x+pi)^2+(y-2pi)^2=8 ?
The solution to (x+pi)^2+(y-2pi)^2=8 is Circle with (a,b)=(-pi,2pi),r=sqrt(8)