Question

How do you find the center, radius, x- and y- intercepts and graph of the circle : x^2+y^2-2x-4y-4=0?

Answer 1

You convert the equation to standard form and use the values of `h` and `k` to calculate these values.

Explanation:

Step 1. Convert the equation to standard form.

The standard form for the equation is `(x-h)^2 + (y-k)^2 = r^2`.

We make the conversion by "completing the square".

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

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

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

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

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

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

`(x-1)^2 + (y-2)^2 = 9`

Step 2. Locate the centre.

The centre is at (`h,k`) = (`x,y`) = (`1,2`).

Step 3. Calculate the radius.

The radius is `r = sqrt9 = 3`.

Step 4. Calculate the `y`-intercepts.

`(x-1)^2 + (y-2)^2 = 9`

Set `x = 0`.

`(0-1)^2 + (y-2)^2 = 9`

`1+ (y-2)^2 = 9`

`(y-2)^2 = 9 -1 = 8`

`y-2 = ±sqrt8 = ±2sqrt2`

`y = 2±2sqrt2`

The `y`-intercepts are at (`0,2-sqrt2`) and (`0,2+sqrt2`).

Step 4. Calculate the `x`-intercepts.

`(x-1)^2 + (y-2)^2 = 9`

Set `y = 0`.

`(x-1)^2 + (0-2)^2 = 9`

`(x-1)^2 + 4 = 9`

`(x-1)^2 = 9 -4`

`(x-1)^2 = 5`

`x-1 = ±sqrt5`

`x = 1 ±sqrt5`

The x-intercepts are at (`1-sqrt5,0`) and (`1+sqrt5,0`).