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

The solution to (x+1)^2+(y-3)^2-2=x+y-2 is Circle with (a,b)=(-1/2 , 7/2),r=sqrt(5/2)

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

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

(a, b) = (- \frac{1}{2}, \frac{7}{2}), r = \frac{5}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - \frac{7}{2})^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, \frac{7}{2}), r = \frac{5}{2}

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

(x + 1)^{2} + (y - 1)^{2} = 36

x^{2} + y^{2} + 6 x + 10 y + 9 = 0

x^{2} + y^{2} - \frac{2}{3} x + \frac{8}{3} y = \frac{19}{9}

x^{2} + 14 x + y^{2} - 2 y - 50 = 0

What is (x+1)^2+(y-3)^2-2=x+y-2 ?
The solution to (x+1)^2+(y-3)^2-2=x+y-2 is Circle with (a,b)=(-1/2 , 7/2),r=sqrt(5/2)