# Teaching tangents and circles at GCSE

Thursday 9th December 2021

When you're introducing how to find the equation of a tangent when given a point on a circle, centre the origin, do you go through the method and give students lots of examples to work out and then move on? Would it be preferable to give an understanding of why the equation of the tangent has such obvious connections to the given point and the given circle? Would you have time to do this and would you make time if you decide it’s important for your Higher GCSE students to investigate this?

Consider this question: Find the equation of the tangent to the circle \(x^2+y^2=25\) at the point \((3,4)\).

A sketch, like the one above, is helpful. Then the process followed is:

- Write down the gradient of the radius: \(m_{r}=\frac{4}{3}\).
- Write down the gradient of the tangent: \(m_{t}=-\frac{3}{4}\).
- Substitute the gradient of the tangent and coordinates of the point into the equation of a straight line to find \(c\): \(y=mx+c\). \(4=-\frac{3}{4}(3)+c\). \(c=\frac{25}{4}\).
- Write down the equation of the tangent and rearrange: \(y=-\frac{3}{4}x+\frac{25}{4}\). \(3x=4y=25\).

After just a few examples, your students may come up with the conjecture that the coefficients of \(x\) and \(y\) in the equation of the tangent match the \(x\) and \(y\) coordinates of the point on the circle, and that the constant value in the equation of the tangent is just the radius squared.

I'm not proposing that students just use this rule instead of showing full working, but I do think it gives an opportunity for students to try and show that this is true for all circles centred at the origin.

Here's another question to consider: Find the equation of a tangent to a circle \(x^2+y^2=r^2\) at the general point on a circle \((p,q)\).

Following the above method:

- \(m_{r}=\frac{q}{p}\)
- \(m_{t}=-\frac{p}{q}\)
- \(y=mx+c\)
- \(q=-\frac{p}{q}(p)+c\)
- \(c=\frac{p^2}{q}+q\)
- \(y=-\frac{p}{q}x+(\frac{p^2}{q}+q)\)
- \(px+qy=p^2+q^2\)

Remember that in this case \(p^2+q^2=r^2\), so the equation of the tangent is \(px+qy=r^2\) for any circle with centre at \((0,0)\).

What if the circle had an equation \((x-a)^2+(y-b)^2=r^2\)?

This would now be going into A level work with circles of centre \((a, b)\), so students have plenty of time to consider this.

By **Jean Smith**