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

Nov 27, 2017

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

Solving simultaneously:

$\setminus \setminus \setminus x + y = 6$
${x}^{2} + {y}^{2} = 18$

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

${x}^{2} + {\left(6 - x\right)}^{2} = 18$

$\therefore {x}^{2} + 36 - 12 x + {x}^{2} = 18$

$\therefore 2 {x}^{2} - 12 x + 18 = 0$

$\therefore {x}^{2} - 6 x + 9 = 0$

${\left(x - 3\right)}^{2} = 0 \implies x = 3$, a repeated root

And with $x = 3$, we have:

$y = 6 - - 3 = 3$

So the line touches the curve at $\left(3 , 3\right)$ 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]}

Nov 27, 2017

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 $\left(3 , 3\right)$, 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.

$\left\{\begin{matrix}{x}^{2} + {y}^{2} = 18 \\ x + y = 6\end{matrix}\right.$

Solving:

$y = 6 - x$

Substituting:

${x}^{2} + {\left(6 - x\right)}^{2} = 18$

${x}^{2} + 36 - 12 x + {x}^{2} = 18$

$2 {x}^{2} - 12 x + 18 = 0$

${x}^{2} - 6 x + 9 = 0$

$\left(x - 3\right) \left(x - 3\right) = 0$

$x = 3$

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!