Show that line $kx - y + 2 - k = 0$ is NOT tangent to the circle $x^2 + y^2 = 9$ for any value of $k$

Question

I have recently been learning about quadratic theory, and I have been given a very difficult question. I must show that line $kx - y + 2 - k = 0$ is NOT tangent to the circle $x^2 + y^2 = 9$ for any value of $k$ using only quadratic theory.

If you have a solution, please make it as explanatory as possible in order to help deepen my understanding further regarding this complicated topic and how you came about with a solution.

Thank you.

Answer 1

The equation of a tangent in $(x_0, y_0)$ to the circle is $xx_0+yy_0=9$.

Now suppose there is such $k$. Then:

$$\frac {k}{x_0} =\frac {-1}{y_0}=\frac {2-k}{-9}$$ and from here: $$\frac {k^2}{(x_0)^2} =\frac {1}{(y_0)^2}=\frac {(2-k)^2}{81}=\frac {k^2 + 1}{(x_0)^2 + (y_0)^2}=\frac {k^2 + 1}{9}$$ therefore $$\frac {(2-k)^2}{81}=\frac {k^2 + 1}{9}$$ with no real solution.

Answer 2

"First line is tangent to the second" equals to "this lines have only one intersection point"

Result of subtitution of $y=kx + 2 -k$ to $x^2 +y^2 = 9$ is

$(k^2+1)x^2 + 2k(2-k)x + (k-2)^2-9 = 0$ This equation (over x) has one root =>

$(k(2-k))^2-(k^2+1)((k-2)^2-9) = 0$

Try to solve this equation over k and check its roots manually

Answer 3

The square of the distance from the line to the origin is clearly $\frac{(2-k)^2}{1+k^2}$. It's easy to show by completing the square that it is at most $5$, so the claim holds even for circles with radius of more than $\sqrt5$.