Dear Uncle Colin,

I’m told there are two circles that touch the x-axis at the origin and are also tangent to the line $4x-3y+24=0$, but I can’t find their equations. Any ideas?

- A Geometrically Nasty Example Seems Impossible

Hi, AGNESI, and thanks for your message!

### Algebraically

If the circle touches the $x$-axis at the origin, its equation must be $x^2 + (y-r)^2 = r^2$, for some value of $k$ - the centre has to be on the $y$-axis, a distance of $r$ away.

You can then play the simultaneous equations game:

• $x^2 + y^2 - 2yr = 0$
• $4x - 3y + 24 = 0$

My approach is to multiply the first equation by 16 and rearrange:

• $(4x)^2 = 32yr - 16y^2$
• $4x = 3y - 24$ so $(4x)^2 = 9y^2 - 144y + 576 = 0$

So $9y^2 - 144y + 576 = 32yr - 16y^2$

Or $25y^2 - (144+32r)y + 576 = 0$

If the circle just touches the line, this system of equations must have a repeated solution - so $(144 + 32r)^2 = 4\times 25 \times 576$.

I could work out $4\times 25 \times 576$, but why would I? They’re all squares.

$144 + 32r = \pm 2\times 5 \times 24 = \pm 240$

$32r = -144 \pm 240$

$r = 3$ or $r=-12$.

So the two circles are $x^2 + (y-3)^2 = 3^2$ and $x^2 + (y+12)^2 = 12^2$ - and that’s the sort of plodding, pedestrian approach you’re supposed to take at A-level.

### Geometrically

The centre of each circle lies on an angle bisector between the $x$-axis and the given line.

The gradient of the given line is $\frac{4}{3}$, and anyone who’s played with a Catriona Shearer problem knows that if you bisect $\arctan\left(\frac{4}{3}\right)$, you get $\arctan\left(\frac{1}{2}\right)$.

So, the upper centre lies on the line with gradient $\frac{1}{2}$ through $(-6,0)$, which crosses the $y$-axis at $(0,3)$, which is the centre of the upper circle.

The lower circle lies on a line perpendicular to that, with gradient $-2$. That crosses the $y$-axis at $(0,-12)$. Job done.

There may be even cleverer ways to do it - I’d love to hear of any!

Hope that helps,

- Uncle Colin

• Thanks to @redmanwinoshoes for the original problem.