Proving a trivial theorem involving a circle and a line

Question

I want to prove that if the length of a line from the origin to the line $Ax + By + C$ is less than the radius of a circle with center $(0, 0)$, then the line $Ax + By + C = 0$ intersects the circle $x^2 + y^2 = r^2$ at two points. I want to prove it because just drawing a graph isnt enough.

To tackle this problem, I used the formula to find the length between a point and a line. Assume the length is $d$ and $d < r$, then $C^2 < r^2(A^2 + B^2)$. This is when $d < r$.

Note that a line $y = mx + n$ intersects a circle with center $(0, 0)$ twice if the discriminant of $x^2 + (mx + n)^2 - r^2 = 0$ is greater than zero. Then I found out that $C^2 < r^2(1+m^2)$ or substituting $m=\frac{-A}B$, I got $C^2 < r^2(\frac{A^2 + B^2}{B^2})$.

My question is how do I go on about using $C^2 < r^2(A^2 + B^2)$ and $C^2 < r^2(\frac{A^2 + B^2}{B^2})$ to prove this 'theorem'?

Answer 1

@Narasiman's answer is the best way. Note that upon solving, if you get real $y$, $x$ is real as well.

Put $x = (-By-C)/A$ in the circle to get: $$\color{red}{(A^2+B^2)}y^2 + \color{green}{2BC}y + \color{blue}{(C^2 - A^2r^2)} = 0$$ $$\implies y = \frac{\color{green}{-2BC} \pm \sqrt{4\color{green}{B^2C^2} + 4\color{blue}{(A^2r^2-C^2)}\color{red}{(A^2+B^2)}}}{\color{red}{2(A^2+B^2)}}$$ $$\implies y = \frac{\color{green}{-BC} \pm \sqrt{\color{green}{B^2C^2} + (\color{orange}{A^2}\color{purple}{r^2}-\color{green}{C^2)}(A^2+B^2)}}{\color{red}{A^2+B^2}}$$ $$\implies y = \frac{\color{green}{-BC} \pm \sqrt{\color{green}{B^2C^2} + \color{orange}{A^2}\cdot \color{purple}{r^2(A^2+B^2)} - \color{orange}{A^2}\color{purple}{C^2} -\color{green}{B^2C^2}}}{\color{red}{A^2+B^2}}$$ $$\implies y = \frac{\color{green}{-BC} \pm \color{orange}A\sqrt{\color{purple}{r^2(A^2+B^2) - C^2}}}{\color{red}{A^2+B^2}}$$

So, $y$ is real iff the purple expression in the square root is non-negative. If the expression is strictly positive, there are 2 values, if the expression is 0, only one value (it becomes a tangent).

In your case since $r^2(A^2+B^2)>C^2 \iff r^2(A^2+B^2)-C^2>0$, $y$ has two values, and so $x$ also has two values. Thus, the line intersects the circle at 2 points.

Answer 2

Solve the two equations

$$ Ax + By + C = 0,~x^2 + y^2 = r^2$$ eliminating either x or y.

The resulting quadratic has two roots, in the form $ a+\sqrt{\Delta}$ for two intersections. The quantity under radical is the discriminant ${\Delta}$.

If discriminant ${\Delta}>0$, there are two real roots or two intersections between the circle and straight line, so we can decide without a sketch or a graph. $d_{min}<r$.

If ${\Delta}=0$, the two roots coincide at a tangent point. $d_{min}=r$.

If $ {\Delta}<0$, the roots are complex and they stay apart without intersection. $d_{min}>r$.