Question

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

Answer 1

`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=mx+6` and your circle equation, `x^2+y^2=9`, as a circle with diameter `3`. So, the two solutions for the line `y=mx+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 `AB` and `AC` are really the two solutions. Therefore, our main objective is to solve for the lengths of the segments `AO` and `BO`; the slopes would be `+-(AO)/(BO)`.

Here's what we know:

`AO=6`, `OP=3`, `AB_|_OP` and `AO_|_BO`

Let's start:

`:'cos(/_AOP)=(PO)/(AO)=1/2`

I use degrees instead of radians here.

`:./_AOP=60°` as `cos(60)=1/2`

Using:

`cos(/_POB)=(OP)/(OB)`

where `/_POB=90-/_AOP=30` and `OP=3`

`cos(30)=3/(OB)`

`OB=3/(cos(30))=2sqrt(3)`

Since slopes are `+-(AO)/(BO)` where `AO=6 and BO=2sqrt(3)`

`+-(AO)/(BO)=+-6/(2sqrt3)=+-sqrt(3)`

Therefore the values of `m` such that `y=mx+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:
`(r, theta)` 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: `(3, theta)`.

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(x)=tan(theta)x`

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

where `b=y+cot(theta)x` for a given point `(x, y)`

We know that `f(x)` and `g(x)` must intersect the circle at `(3, theta)`, but these are polar coordinates. To get the cartesian coordinates, we use:
`y=rsintheta` and x=`rcostheta`

We plug it in to find `b`:

`b=rsintheta+rcostheta*cottheta`
`r=3`

Thus `g(x)=-cot(theta)x+3sintheta+3costheta*cottheta`

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

So solve for `theta` in `rsintheta+rcostheta*cottheta=6` and we get, in radians this time:
`pi/6` and `5/6pi` between `[0, 2pi)`

We plug it back into `-cot(theta)` to give us the slopes.

Told you it was overly complex.