Chapter 4 · Geometry and Equations · Concept 3
4-22 Length of a Chord
A line that cuts through a circle leaves a piece of itself inside — the chord. You could find both ends and measure between them, but there is a far quicker way: drop a perpendicular from the centre and a right-angled triangle appears, with the radius as its longest side.
Guiding question
How wide is the circle where the road crosses it?
Learning outcome. Compute the length of the chord formed where a line crosses a circle.
Draw the diagram, then use Pythagoras
r2 = d2 + l2
r — the radius of the circle
d — the perpendicular distance from the centre down to the chord
l — half the length of the chord
Why half, and why a right angle.
A perpendicular dropped from the centre of a circle to a chord always bisects it — that is the old theorem from Math A, and it is what makes the triangle work. The two halves are mirror images, so the triangle formed by the radius, the perpendicular and half the chord has a right angle at the foot.
l is half. Double it at the end.
Pythagoras gives you l, and l is not the answer — the chord is 2l. And when the chord is given to you instead, halve it before it goes into the triangle. This one factor of 2 accounts for more lost marks on this page than everything else put together.
d comes from lesson 4-12.
Nothing new: d = |ax₁ + by₁ + c| ÷ √(a² + b²), with the centre as the point. Tidy the line into ax + by + c = 0 form first, every time.
The book’s first instruction is “draw a diagram”, and it means it. Once the triangle is on the page the algebra is three lines.
x² + y² = 4 · x + y − 1 = 0
1. Find the length of the chord cut off by x² + y² = 4 on the line x + y − 1 = 0.
- 1
Read the circle. Centre and radius first, always.
centre (0, 0), r = 2x² + y² = 4 is centred at the origin with r² = 4, so r = 2.
Work with a partner. Sketch question 2 roughly before solving it. Two positions of the line give the same chord — one on each side of the centre — and seeing that on paper is why the answer has a ± in it.
1–3. Find the chord.
4–5. The chord is given; find k.
Question 5 does not give ±k, and that is worth noticing.
In question 4 the centre is the origin, so the absolute value is |k| and the two answers are mirror images. In question 5 the centre is (1, 0), so it is |1 + k| = 1 — which opens to 1 + k = 1 or 1 + k = −1, giving k = 0 and k = −2. Still two answers, but no longer a matching pair. Never assume the ± before you have written the absolute value out.
Find the value of the constant k such that the line y = x + k cuts the circle x² + y² + 4x − 2y − 7 = 0 in a chord of length 4.
Giving l as the chord.
Pythagoras hands you HALF. The answer is 2l, and this is by far the most common error on the page.
Feeding the whole chord into the triangle.
Going backwards, halve the given chord first. A chord of 4 means l = 2, not 4.
Assuming the answer is ±k.
It is only ± when the centre is the origin. With centre (1, 0) it is |1 + k| = 1, giving k = 0 or −2 — not a matching pair.
Using the triangle on a general-form circle.
x² + y² + 4x − 2y − 7 = 0 has to be completed into standard form first. Without a centre and a radius there is no triangle.
Taking square roots too early.
Stay in d² and l² for as long as possible. The surds appear once, at the very end, instead of being dragged through every line.
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