## A lovely circle problem – two ways

So, I was working with some year 12s on a few problems around circles out of the new Pearson A Level textbook. (Incidentally, it’s this book, and I think it’s probably the best textbook I’ve come across. I would certainly recommend it.)

This question appears in a mixed exercise on circles:

It’s a lovely question. Before reading on, have a go at it – or at least have a think about what approach you’d take – as I’m going to discuss a couple of methods and I’d be interested to know how everyone else approached it.

Method 1:

I looked at this problem and saw right angled triangles with the hypotenuse root 52. I knew the gradient of the radii must be -2/3 as each radius met a tangent of radius gradient 3/2. From there it followed logically that the ratio of vertical side : horizontal side is 2 : 3.

Using this I could call the vertical side 2k and the horizontal side 3k. Pythagoras’s Theorem then gives 13k^2 = 52, which leads to k^2 is 4 and then k is 2 (or -2).

So the magnitude of the vertical side is 4 and of the horizontal side is 6.

From here it follows nicely that p is (-3,1) and q is (9, -7).

Finally there was just the case working out the equation given a gradient and a point.

L1: y – 1 = (3/2)(x +3)

2y – 2 = 3x + 9

3x – 2y + 11 = 0

L2: y + 7 = (3/2)(x -9)

2y + 14 = 3x – 27

3x – 2y – 41 = 0

I thought this was a lovely solution, but it seemed like a rather small amount of work for an 8 mark question. This made me wonder what the marks would be for, and then it occurred to me that perhaps this wasn’t the method the question writer had planned. Perhaps they had anticipated a more algebraic approach.

Method 2:

I had the equation of a circle: (x – 3)^2 + (y + 3)^2 = 52. I also knew that each tangent had the equation y = (3/2)x + c. It follows that if I solve these simultaneously I will end up with a quadratic that has coefficients and constants in terms of c. As the lines are tangents, I need the solution to be equal roots, so by setting the discriminant equal to zero I should get a quadratic in c which will solve to give me my 2 y intercepts. Here are the photos of my workings.

As you can see, this leads to the same answer, but took a lot more work.

*I’d love to know how you, or your students, would tackle this problem.*

I like your first observation about the triangles. My first method was to just note the equation of the radial line was y = -2/3x +b and goes through 3,-3. Substitute the point in and b must be -1.

Then you can just plug y = -2/3x -1 into the circle equation to find the tangent points

i.e. (x -3)^2 + ((-2/3x -1) + 3)^2 = 52 and out pops 9 and -3 from which you get the points (9,7) (-3,1)

Finally just plug those points into the equation y = 3/2x + b to find the final lines. (and normalize to meet the required equation form).

So very much like your 2nd approach but its simpler if you solve for the c first.

That’s a very nice alternative. Thanks.

This was how I approached it too. Which is great because it’s saved me typing it out!