Question

How do you find the center and radius of the circle `x^2-2x+y^2-8y+1=0`?

Answer 1

Radius: `4`
Centre: `(-1, 4)`

Explanation:

You should convert to the form `(x - a)^2 + (y- b)^2 = r^2`. Complete the square twice to achieve this.

`1(x^2 - 2x + n) + 1(y^2 - 8y + m) = -1`

Our goal here is to make the expressions in parentheses into perfect squares. We can use the formula `n = (b/2)^2` and `m = (b/2)^2`, where `b` is the x-coefficent term.

So,

`n = (-2/2)^2 = 1`

`m = (-8/2)^2 = 16`

We need to add and subtract this number inside the parentheses to keep the equation equivalent.

`1(x^2 - 2x + 1 - 1) + 1(y^2 - 8x + 16 - 16) = -1`

`1(x^2 - 2x + 1) - 1 + 1(y^2 - 8x + 16) - 16 = -1`

`(x - 1)^2 + (y - 4)^2 = -1 + 17`

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

The radius of a circle in the form `(x - a)^2 + (y - b)^2 = r^2` is `r`, so in this case `4`. The centre is at `(a, b)`, in this case `(-1, 4)`.

Hopefully this helps!

Answer 2

You complete the squares by adding `h^2 and k^2` to both sides of the equation. Please see the explanation.

Explanation:

The standard Cartesian form for the equation of a circle is:

`(x - h)^2 + (y - k)^2 = r^2" [1]"`

where `(x, y)` is any point on the circle, `(h, k)` is the center, and `r` is the radius.

Equation [2] is the same as equation [1] but with the squares expanded and equation [3] is the given equation with some spaces added for missing terms:

`x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"`
`x^2 - 2x " " + y^2 - 8y" "+ 1= 0" [3]"`

Let's try to make equations [2] and [3] match by adding `h^2 and k^2` to both sides of equation [3] and label it equation [4]:

`x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"`
`x^2 - 2x + h^2 + y^2 - 8y + k^2 + 1= h^2 + k^2" [4]"`

Subtract 1 from both sides of equation [4] and label it equation [5]:

`x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"`
`x^2 - 2x + h^2 + y^2 - 8y + k^2 = h^2 + k^2 - 1" [5]"`

Now that equation [2] and equation [5] both have 6 terms on the left side, we can see that we can find the value of h by equating the second term in equation [2] with the same term in equation [5]:

`-2hx = -2x`

`h = 1`

Substitute 1 for h into equation [5] and number it equation [6]:

`x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"`
`x^2 - 2x + 1^2 + y^2 - 8y + k^2 = 1^2 + k^2 - 1" [6]"`

We can find the value of k by equating the fifth term in equation [2] with the same term in equation [5]:

`-2ky = -8y`

`k = 4`

Substitute 4 for k into equation [6] and number it equation [7]:

`x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [2]"`
`x^2 - 2x + 1^2 + y^2 - 8y + 4^2 = 1^2 + 4^2 - 1" [7]"`

Because we used the patterns for `(x - h)^2 and (y - k)^2`, respectively, we know that the first three terms become `(x - 1)^2` and the next three terms become `(y - 4)^2`:

`(x - 1)^2 + (y - 4)^2 = 1^2 + 4^2 - 1" [8]"`

Simplify the right side:

`(x - 1)^2 + (y - 4)^2 = 4^2" [9]"`
`(x - h)^2 + (y - k)^2 = r^2" [1]"`

Comparing equation [9] with equation [1], we can see that the center is `(1, 4)` and the radius is 4.