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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