# Ask Uncle Colin: A Couple of Tangents

Dear Uncle Colin,

I’ve got two circles ($x^2 + y^2 = 3^2$ and $(x-10)^2 + y^2 =5^2$) and I want to find the equations of the common tangents. I’ve been stuck for ages!

- Tackling A New Geometry Exercise, Need Tuition

Hi, TANGENT, and thanks for your message!

I can see two fairly nice ways to approach this.

### With algebra!

Suppose the equation of a tangent line is $y = m(x-x_0)$ for some fixed values of $m$ and $x_0$. ((Other equations are available.))

This line meets the first circle when $x^2 + m^2(x-x_0)^2 = 3^2$; if the line is tangent, the equation has a repeated root.

Rearrange to get $x^2 (1+m^2) - 2m^2x_0x + (m^2x_0^2 - 9) = 0$; this has equal roots if $4m^4x_0^2 = 4(1+m^2)(m^2 x_0^2 - 9)$.

Expanding that and dividing by 4 gives $m^4 x_0^2 = m^2x_0^2 - 9 + m^4x_0^2 - 9m^2$

There’s an $m^2 x_0^2$ on each side, so $m^2(x_0^2 - 9) = 9$, or $m^2 = \frac{9}{x_0^2 - 9}$.

So, this is the relationship between $m$ and $x_0$ for any line that touches the first circle. What about the second one?

We can do a similar analysis. Substitute to get $(x-10)^2 + m^2(x-x_0)^2 = 5^2$, and rearrange to get $x^2 (1+m^2) - (2m^2 x_0 + 20)x + (75 + m^2x_0^2)= 0$.

Next, this needs to have repeated roots, so $(2m^2x_0 + 20)^2 = 4(1+m^2)(75 + m^2x_0^2)$.

Expanding and dividing by 4: $m^4 x_0^2 + 20m^2 x_0 x + 100 = 75 + m^2x_0^2 + 75m^2 + m^4x_0^2$.

Again, the $m^4$ term vanishes to leave us, this time, with $m^2(x_0^2-20x_0 + 75) = 25$.

We have two relationships now:

- $m^2 = \frac{9}{x_0^2 - 9}$
- $m^2 = \frac{25}{x_0^2 - 20x_0 + 75}$

We can combine those to give $\frac{9}{x_0^2 - 9} = \frac{25}{x_0^2 - 20x_0 + 75}$, or $9(x_0^2 - 20x_0 + 75) = 25(x_0^2 - 9)$.

This rearranges to $0 = 16x_0^2 + 180x_0 - 900$, or $4x_0^2 + 45x_0 - 225 = 0$…

**whoosh**

“$(4x_0-15)(x_0+15)$. You’re welcome.”

**whoosh**

So $x_0 = \frac{15}{4}$ or $-15$, and $m^2 = \frac{16}{9}$ or $\frac{1}{24}$.

The tangent lines are $y = \pm \frac{4}{3}\left( x - \frac{15}{4}\right)$ and $y = \pm \frac{1}{\sqrt{24}}\left( x + 15 \right)$.

The Mathematical Ninja, I couldn’t help noticing, wrinkled their face in passing.

### With geometry:

We have a circle with radius 3, centred at the origin.

We have a circle with radius 5, centred at point P, (-10, 0).

Suppose we have a line that’s tangent to the first circle at T and the second circle at U, and that crosses the X-axis at X.

Triangle OTX and triangle PUX are similar, with a scale factor of 3:5.

Now, there are two classes of tangent: inner tangents (that cross between the circles) and outer tangents (that cross outside). Let’s look at those in turn.

If the tangents cross the axis outside the circles, we know that the difference between the hypotenuses of the triangles is 10. That means the longer hypotenuse is 25 units long and the shorter one 15 – and these must meet the axis at (-15,0).

If the shorter hypotenuse is 15 and its shorter leg is 3, its longer leg is $\sqrt{225-9} = 6\sqrt{6}$. The gradient of the tangent is then $\pm \frac{3}{6\sqrt{6}} = \pm \frac{1}{\sqrt{24}}$, so the equation of the line is $y = \pm \frac{1}{\sqrt{24}}(x+15)$. (The $\pm$ is because the tangent could be above or below the axis).

Looking at the inner tangents, the sum of the hypotenuses is 10, so their lengths are $\frac{15}{4}$ and $\frac{25}{4}$, and they cross at $\left(\frac{15}{4},0\right)$.

The shorter hypotenuse is $\frac{15}{4}$ long, and one of the legs is 3, so the other is $\frac{9}{4}$. The gradient of the tangent is $\pm\frac{9}{3/4} = \pm\frac{4}{3}$, and the equation of the line is $y = \frac{4}{3}\left(x - \frac{15}{4}\right)$.

Hope that helps!

- Uncle Colin