Chapter 4 · Geometry and Equations · Concept 3
4-17 Equation of a Circle (2)
This time nobody hands you the centre and the radius. The question describes the circle in a roundabout way — a point it passes through, the two ends of a diameter, or three scattered points — and your first job is to turn that description into the two numbers you actually need.
Guiding question
Three dots on a plate. Can you find the plate?
Learning outcome. Find the equation of a circle from a point on it, from a diameter, or from three points.
Three descriptions, three first moves
given a CENTRE and a POINT on it
The radius is just the distance between them. Find r2 with the distance formula and stop — you never need r itself.
given the two ends of a DIAMETER
The centre is the midpoint of those two ends. Then the radius is the distance from that centre out to either end.
given THREE POINTS on it
Use the general form x² + y² + lx + my + n = 0. Put each point in to get three equations, then solve for l, m and n.
For three points, use the general form — not the standard one.
In standard form the unknowns a and b sit inside squares, so substituting gives you a2 and b2 terms and the equations turn nasty. In general form the unknowns l, m and n are all plain and straight — three ordinary simultaneous equations. Same circle, far less pain.
Every question of the third kind gives you exactly three points. Never two, never four. This is why.
Two points on a circle. That sounds like plenty — but watch what happens.
3. Find the circle passing through (1, 1), (2, −1) and (3, 2).
- 1
Start from the general form, with three things to find.
x2 + y2 + lx + my + n = 0l, m and n are the unknowns. Three unknowns need three facts — and we have exactly three points.
Always test the third point.
Put (3, 2) into x² + y² − 5x − y + 4: 9 + 4 − 15 − 2 + 4 = 0. It works. That one line catches almost every arithmetic slip in the whole page.
Work with a partner. For the three-point question, one of you solves it and the other checks by substituting all three points back into the finished equation. All three must give exactly 0 — if even one does not, the answer is wrong, no matter how tidy the working looked.
1–4. A centre and a point on the circle.
5–8. The ends of a diameter.
9–11. Three points.
Look for the friendly point first.
In 9 one of the points is the origin, which gives you n = 0 immediately and turns three unknowns into two. In 10 the point (0, 8) has x = 0, so l vanishes from that equation. Starting with the point that has a zero in it saves a page of work every time.
A circular sensor ring is being mapped. Three sensors on the ring report their positions as (0, 2), (2, 0) and (4, 6). Find the equation of the ring, and state where its centre is and how far it reaches.
Treating a diameter end as the centre.
The two given points are on the RIM. The centre is halfway between them — take the midpoint before doing anything else.
Using the diameter length as the radius.
The distance between the two ends is the whole diameter. The radius is half of it, or the distance from the midpoint out to one end.
Trying three points in standard form.
Substituting into (x − a)² + (y − b)² = r² gives you a², b² and r² all at once. Use the general form, where l, m and n are plain.
Sign slips when a coordinate is negative.
Putting (2, −1) into my gives −m, not m. Most wrong answers on this page trace back to exactly one of these.
Never checking the answer.
All three points must give exactly 0 in the finished equation. It takes seconds and it catches almost everything.
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