# Ask Uncle Colin: Tangents to a circle

Dear Uncle Colin,

I'm told that two lines through $(0,12)$ are tangent to the circle with equation $(x-6)^2 + (y-5)^2 = 17$ and I need to find their equations - but I'm getting in a muddle. Can you help?

- Terribly Awkward Numbers, Getting Equations Not Trivial

Hi, TANGENT, and thank you for your message!

This certainly is a non-trivial one, and the numbers are definitely awkward!

The main difficulty for most people here is keeping a clear distinction between variables - such as the $x$ and $y$ that represent points in space - and parameters - such as $m$, which represents the gradient of a line.

Every possible line through $(0,12)$ can be written as $y= mx + 12$ for some specific value of the gradient parameter $m$. Given $m$, you can check whether any point in space lies on the line by simply putting the values into each side of the equation and seeing whether it holds true - supposing $m=2$, you can see that $(1,14)$ is on the line because the left hand side and right hand side both evaluate to 14. However, $(9,7)$ doesn't lie on the line, because the left hand side works out to be 7 and the right hand side 30.

Now, the first line of attack is to find out where any such line meets the circle.

### Where they meet

To do this, we substitute for $y$ in the circle equation to get: $(x-6)^2 + (mx+7)^2 = 17$.

When this is fully expanded and simplified, we find that (for a given $m$), the line meets the circle wherever $\br{1+m^2} x^2 + (14m-12)x + 68 = 0$.

Depending on the value of $m$, this line either cuts the circle in two places, misses it altogether, or just grazes it, giving a repeated root. In the last of these cases, the line is a tangent.

### Repeated root

For the quadratic equation to have a repeated root, its discriminant must be zero. That means, $(14m-12)^2 = 4 \times 68 (1+m^2)$.

Expanding this gives $196m^2 - 168m + 144 = 272 + 272m^2$, or $0=76m^2 + 168m - 128$. Dividing by 4 leads to $0 = 19m^2 + 42m + 32$.

Why do we have a quadratic in $m$? Well, each value of $m$ corresponds to a line, and two of those must be tangents - so we should really expect to have an equation with two solutions!

### Does it factorise?

You betcha. It's $(19m + 8)(m + 4)=0$, although you presumably have the kind of calculator that does that for you.

That gives the two possible gradients of $-4$ and $-\frac{8}{19}$.

The two possible equations of the line are $y = -4x + 12$ and $y=-\frac{8}{19}x + 12$.

Hope that helps!

- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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