Question

How do you find the standard form given `x^2+y^2-8x+4y+11=0`?

Answer 1

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

Explanation:

Firstly, which conic is this?

Well, the coefficients in front of the `x^2` and `y^2` terms are the same, and there is a plus sign between them, so it is a circle.

The standard form equation of a circle is given by

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

where

• `h` is the `x`-coordinate of the center of the circle

• `k` is the `y`-coordinate of the center of the circle

• `r` is the radius of the circle

We'll need to complete the square to find the standard form equation.

Let's move the constant `11` to the right hand side, and group `x`- and `y`-terms together:

`x^2 - 8x + y^2 + 4y = -11`

Set up the completion of the square for `x` and `y`:

`(x^2 - 8x + ul(" ")) + (y^2 + 4y + ul(" ")) = -11`

We're looking for the "`c`" term in the parabolic equation, and to find that we do

`ul(c = (b/2)^2`

or

`ul(c = (b^2)/4`

The `b` term for the `x` is `-8` (the variable in front of the linear term `x`), so we have

`x{color(white)(a)c = ((-8)^2)/4 = color(red)(ul(16`

Similarly, for the `y`, `b = 4`, so we have

`y{color(white)(a)c = (4^2)/4 = color(green)(ul(4`

Now, we plug these values into the equation:

`(x^2 - 8x + color(red)(16)) + (y^2 + 4y + color(green)(4)) = -11`

Now, we add both values of `c` to the constant on the right side:

`(x^2 - 8x + color(red)(16)) + (y^2 + 4y + color(green)(4)) = -11 + color(red)(16) + color(green)(4)`

Each of the terms can now be represented as the square of a monomial, and so we have

`color(blue)(ulbar(|stackrel(" ")(" "(x-4)^2 + (y+2)^2 = 9" ")|)`

(to get the value of `h` and `k`, simply take half of the `b` value for each variable; i.e. `overbrace(-8/2 = -4)^x` and `overbrace(4/2 = 2)^y`

For this equation here, we can determine:

• the center point `(h,k)` of the circle is `(4, -2)`

• the radius `r` is `sqrt9 = 3`