Question

At what point(s) does the line `x + y =6` and the circle `x^2 + y^2 = 18` intersect?

Answer 1

The line will be tangent to the curve it it intersects the curve at a repeated solution.

Solving simultaneously:

`\ \ \ x+y=6`
`x^2+y^2 = 18`

Thus by eliminating `y` using `y=6-x` we have:

`x^2 + (6-x)^2 = 18`

`:. x^2 + 36 -12x + x^2 = 18`

`:. 2x^2 -12x + 18 = 0`

`:. x^2 -6x + 9 = 0`

`(x-3)^2 = 0 => x=3`, a repeated root

And with `x=3`, we have:

`y = 6-- 3 = 3`

So the line touches the curve at `(3,3)` and therefore must be a tangent, which we can confirm graphically:

graph{ (x^2+y^2 - 18)(x+y-6)=0 [-10, 10, -5, 5]}

Answer 2

Graphically

We can graph the circle `x^2 + y^2 = 18` and `x + y = 6` on the same grid.

We can see that the line touches the circle at one point, namely `(3, 3)`, and thus we can say that the line `x + y = 6` is tangent to `x^2 + y^2 = 18`.

Algebraically

We solve the following system of equations to check if the two curves indeed intersect.

`{(x^2 + y^2 = 18), (x + y = 6):}`

Solving:

`y = 6 - x`

Substituting:

`x^2 + (6 - x)^2 = 18`

`x^2 + 36 - 12x + x^2 = 18`

`2x^2 - 12x + 18 = 0`

`x^2 - 6x + 9 = 0`

`(x -3)(x - 3) = 0`

`x = 3`

Now solving for `y`, we get `y = 6 - 3 = 3`. Since there is only one point of intersection, we can guarantee that `x^2 + y^2 = 18` and `x + y = 6` are tangent.

Hopefully this helps!