# Error in previous question, what is radius of circle with equation x^2 + y^2 + 2x - 2y - 4 = 0

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`x^2 + y^2 + 2x - 2y - 4 = 0` Find the radius.

The general equation with a circle with center (h, k) and radius, *r*, is:

`(x - h)^2 + (y - k)^2 = r^2`

` `We need to complete the square in order to write the original equation in general equation form.

First: `(x^2 + 2x + a) + (y^2 - 2y + b) = 4`

In order to find *a* and *b* divide middle term of each trinomial by 2 then square it. This gives us:

`(x^2 + 2x + 1) + (y^2 - 2y + 1) = 4 + 1 + 1`

Now factor,

`(x + 1)^2 + (y - 1)^2 = 6`

Since `6 = r^2,`

`r =sqrt(6)`

**Therefore, the radius of the circle is** `sqrt(6).`

The equation of a circle with center (h, k) and radius r is `(x - h)^2 + (y - k)^2 = r^2`

Writing the equation given in this form:

`x^2 + y^2 + 2x - 2y - 4 = 0`

=> `x^2 + 2x + 1 + y^2 - 2y + 1 - 4 - 2 = 0`

=> `(x + 1)^2 + (y - 1)^2 = 6`

=> `(x + 1)^2 + (y - 1)^2 = (sqrt 6)^2`

**The radius of the circle with equation x^2 + y^2 + 2x - 2y - 4 = 0 is **`sqrt 6`