What is the radius x^2+y^2-2x+2y-7=0 ?

The radius x^2+y^2-2x+2y-7=0 is r=3

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radius x^2+y^2-2x+2y-7=0

radius

x^{2} + y^{2} - 2 x + 2 y - 7 = 0

r = 3

Solution steps

x^{2} + y^{2} - 2 x + 2 y - 7 = 0

Rewrite

x^{2} + y^{2} - 2 x + 2 y - 7 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 1), r = 3

And the radius is:

r = 3

x^{2} + y^{2} - 10 x + 7 y - 29 = 0

9 x^{2} + 54 x + 9 y^{2} - 6 y + 28 = 0

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

x^{2} + y^{2} - 8 x + 18 y + 16 = 0

(x - 1)^{2} + (y + 4)^{2} = 16