Question

How do you write `x^2+y^2+8x+7=0` in standard form?

Answer 1

`(x+4)^2 + y^2 = 9`
circle with center `(-4, 0), " "r = 3`

Explanation:

Given: `x^2 + y^2 + 8x + 7 = 0`

You must use completing of the square. First group the `x`-terms together, the `y`-terms together and put the constants on the right side:

`(x^2 + 8x) + y^2 = -7`

To complete the square, half the `8x` term constant = `4` and add the square of this constant (4^2 = 16) to the right side.

We do this because `(x+4)^2 = x^2 + 8x + 16`. When the square is completed there will always be a constant that needs to be added to the other side to keep the equation balanced.

`(x+4)^2 + y^2 = -7 + 16`

`(x+4)^2 + y^2 = 9`

This is the standard form of a circle: `(x - h)^2 + (y-k)^2 = r^2`

where the center is `(h, k)` and the radius `= r`