# For what value(s) of m is the line y=mx+6 tangent to the circle x²+y²=9?

Oct 14, 2017

$\sqrt{3}$ and $- \sqrt{3}$

#### Explanation:

I highly recommend you visualize the equations on a graph - it is very helpful to do so. Imagine the function, $y = m x + 6$ and your circle equation, ${x}^{2} + {y}^{2} = 9$, as a circle with diameter $3$. So, the two solutions for the line $y = m x + 6$ will look something like the orange lines shown in the picture above.

An interesting property of tangent lines is that a line drawn from the point of intersection to the circle's center is always perpendicular to the tangent line.

A way to approach this problem is to exploit this property of perpendicular tangency. Now, originally, I wanted to solve this problem using purely just the equations, however, it became really complex.

However, I realized that this can be done with a bit of simple geometry and a touch of trigonometry, here's how:

Refer to the above diagram. The line segments $A B$ and $A C$ are really the two solutions. Therefore, our main objective is to solve for the lengths of the segments $A O$ and $B O$; the slopes would be $\pm \frac{A O}{B O}$.

Here's what we know:

$A O = 6$, $O P = 3$, $A B \bot O P$ and $A O \bot B O$

Let's start:

$\because \cos \left(\angle A O P\right) = \frac{P O}{A O} = \frac{1}{2}$

I use degrees instead of radians here.

:./_AOP=60° as $\cos \left(60\right) = \frac{1}{2}$

Using:

$\cos \left(\angle P O B\right) = \frac{O P}{O B}$

where $\angle P O B = 90 - \angle A O P = 30$ and $O P = 3$

$\cos \left(30\right) = \frac{3}{O B}$

$O B = \frac{3}{\cos \left(30\right)} = 2 \sqrt{3}$

Since slopes are $\pm \frac{A O}{B O}$ where $A O = 6 \mathmr{and} B O = 2 \sqrt{3}$

$\pm \frac{A O}{B O} = \pm \frac{6}{2 \sqrt{3}} = \pm \sqrt{3}$

Therefore the values of $m$ such that $y = m x + 6$ is tangent to ${x}^{2} + {y}^{2} = 9$ are $\sqrt{3}$ and $- \sqrt{3}$.

But wait there's more

Remember how I said there was an algebraic way to do it? And that it's overly complicated? Here it is (no diagrams):

Any point on this circle can be expressed in polar coordinates:
$\left(r , \theta\right)$ where $r$ is the radius and $\theta$ is the degrees offset from the $x$ axis.
Therefore any point on the circle can be rewritten as: $\left(3 , \theta\right)$.

A tangent line on the circle is perpendicular to the radius drawn from the point of intersection. Let $\theta$ represent the degrees offset of the radius, therefore the radius can be written as this function:
$f \left(x\right) = \tan \left(\theta\right) x$

Then we need to write a function describing the tangent line:
$g \left(x\right) = - \cot \left(\theta\right) x + b$

where $b = y + \cot \left(\theta\right) x$ for a given point $\left(x , y\right)$

We know that $f \left(x\right)$ and $g \left(x\right)$ must intersect the circle at $\left(3 , \theta\right)$, but these are polar coordinates. To get the cartesian coordinates, we use:
$y = r \sin \theta$ and x=$r \cos \theta$

We plug it in to find $b$:

$b = r \sin \theta + r \cos \theta \cdot \cot \theta$
$r = 3$

Thus $g \left(x\right) = - \cot \left(\theta\right) x + 3 \sin \theta + 3 \cos \theta \cdot \cot \theta$

This will give us a tangent line to the circle. However, we have restraints, and that is $b = 6$ (recall the question $y = m x + b$ rewritten as $g \left(x\right) = m x + b$ ).

So solve for $\theta$ in $r \sin \theta + r \cos \theta \cdot \cot \theta = 6$ and we get, in radians this time:
$\frac{\pi}{6}$ and $\frac{5}{6} \pi$ between $\left[0 , 2 \pi\right)$

We plug it back into $- \cot \left(\theta\right)$ to give us the slopes.

Told you it was overly complex.