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S6 Maths | Locus and Circle Equations Lesson-Prep Guide (for tutors) | Perpendicular bisectors, general form, parabolic loci
Students can recite that a locus is the set of points satisfying a condition, but the moment coordinates appear they hunt for a formula, and x squared plus y squared minus 4x plus 6y minus 12 equals zero tells them nothing. This lesson reduces all three conditions to one move: write the equal distances.
Objectives — Students can: (1) translate equidistance from two points into a perpendicular bisector and check it by substitution; (2) convert between the standard and general forms of a circle and read the centre and radius straight from D, E and F; (3) translate equidistance from a point and a line into a parabola and justify the vertex and the opening direction.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Distance between points | Distance from A(1, 2) to B(5, 10) | Root 80, about 8.94 |
| Gradient | Gradient of AB | (10 – 2) over (5 – 1) = 2 |
| Completing the square | Turn x squared minus 4x into a perfect square | (x – 2) squared minus 4 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Definition and translation table | Write four conditions as distance statements | Four equations |
| 8-20 min | Perpendicular bisector | A(1, 2) and B(5, 10) | x + 2y – 15 = 0 |
| 20-34 min | Standard and general form | Centre (2, -3), radius 5 | F = -12 |
| 34-44 min | Three points define a circle | Diameter endpoints and three points | Centre (3, 4), r = 5 |
| 44-54 min | Equidistant from a point and a line | Point (0, 0), line x = 4 | y squared = -8(x – 2) |
| 54-60 min | Exit check | Three questions, marked on the spot | Independent |
03 Teaching script (read it out)
Scene 1: A locus is a distance statement, not a drawing
Equidistant from A and B means exactly PA = PB. Write the distances, square both sides, and the root disappears: the midpoint is (3, 6), the gradient of AB is 2, so the perpendicular gradient is minus one half and the line is x + 2y – 15 = 0.
Checking beats deriving. Substituting A gives minus 10 and B gives plus 10: equal size, opposite sign, which is exactly what equidistant and opposite sides look like.
The same translation handles a fixed distance from a line, but the answer is two parallel lines, not one, and that is the part students drop.
Checking beats deriving. Substituting A gives minus 10 and B gives plus 10: equal size, opposite sign, which is exactly what equidistant and opposite sides look like.
The same translation handles a fixed distance from a line, but the answer is two parallel lines, not one, and that is the part students drop.
Have students write the bisector and then self-check by substitution.
Scene 2: One circle, two ways of writing it
The standard form (x – 2) squared plus (y + 3) squared equals 25 shows the centre (2, -3) and radius 5 at a glance; expanded it becomes x squared plus y squared minus 4x plus 6y minus 12 equals zero.
Going back needs three links: D = -2h, E = -2k, F = h squared plus k squared minus r squared. So the centre is (-D over 2, -E over 2) and the radius is the root of (D over 2) squared plus (E over 2) squared minus F, which is the root of 4 plus 9 plus 12, that is 5.
From diameter endpoints (1, 1) and (7, 9) the midpoint (4, 5) is the centre and half the length, 5, is the radius, giving x squared plus y squared minus 8x minus 10y plus 16 equals zero. There is a faster route: the angle in a semicircle is a right angle, so (x – 1)(x – 7) + (y – 1)(y – 9) = 0.
Going back needs three links: D = -2h, E = -2k, F = h squared plus k squared minus r squared. So the centre is (-D over 2, -E over 2) and the radius is the root of (D over 2) squared plus (E over 2) squared minus F, which is the root of 4 plus 9 plus 12, that is 5.
From diameter endpoints (1, 1) and (7, 9) the midpoint (4, 5) is the centre and half the length, 5, is the radius, giving x squared plus y squared minus 8x minus 10y plus 16 equals zero. There is a faster route: the angle in a semicircle is a right angle, so (x – 1)(x – 7) + (y – 1)(y – 9) = 0.
Have students explain why three points fix a circle but two do not.
Scene 3: Equidistant from a point and from a line: one is straight, one bends
A point equidistant from (0, 0) and from the line x = 4 satisfies root of (x squared plus y squared) equals the modulus of x minus 4. Squaring gives y squared = -8x + 16, that is y squared = -8(x – 2).
The vertex is (2, 0) and it opens leftwards, away from the line, because the moving point sits on the far side of the focus from the directrix. The sign is the usual casualty: y squared = +8(x – 2) can be killed in one line by testing x = 0, which gives y = plus or minus 4, and (0, 4) is 4 from the origin and 4 from x = 4.
Tangency is the other family: the line y = x + k touches x squared plus y squared = 25 when the distance from the centre equals the radius, so |k| over root 2 is 5 and k is plus or minus 7.07. Two answers, one on each side.
The vertex is (2, 0) and it opens leftwards, away from the line, because the moving point sits on the far side of the focus from the directrix. The sign is the usual casualty: y squared = +8(x – 2) can be killed in one line by testing x = 0, which gives y = plus or minus 4, and (0, 4) is 4 from the origin and 4 from x = 4.
Tangency is the other family: the line y = x + k touches x squared plus y squared = 25 when the distance from the centre equals the radius, so |k| over root 2 is 5 and k is plus or minus 7.07. Two answers, one on each side.
Insist that every locus answer is tested with one concrete point.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Perpendicular bisector | Midpoint, negative reciprocal, point-slope form | x + 2y – 15 = 0 |
| Calculation | Centre and radius from general form | Centre (-D/2, -E/2), radius from the three coefficients | (2, -3), r = 5 |
| Calculation | Circle from three points or a diameter | Midpoint as centre, half length as radius | x squared + y squared – 6x – 8y = 0 |
| Explanation | Classify a locus | Write the distance relation, then look at the degree | Straight line or parabola |
Sanity check (teach this): The radius squared must be positive: 25 is fine, a negative means F is too large or D and E were miscopied.
The bisector must give equal magnitudes with opposite signs: minus 10 and plus 10 agree, matching signs mean the reciprocal sign was dropped.
Tangency always gives two values of k: plus and minus 7.07; one answer means half the picture is missing.
The parabola coefficient must agree with the vertex: vertex (2, 0), coefficient -8, opening left; a positive coefficient with a leftward claim is self-contradictory.
Every locus answer survives one substitution, and that check needs no formula at all.
The bisector must give equal magnitudes with opposite signs: minus 10 and plus 10 agree, matching signs mean the reciprocal sign was dropped.
Tangency always gives two values of k: plus and minus 7.07; one answer means half the picture is missing.
The parabola coefficient must agree with the vertex: vertex (2, 0), coefficient -8, opening left; a positive coefficient with a leftward claim is self-contradictory.
Every locus answer survives one substitution, and that check needs no formula at all.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Locus equidistant from (1, 2) and (5, 10) | x + 2y – 15 = 0 | Midpoint and negative reciprocal |
| Centre and radius of x squared + y squared – 4x + 6y – 12 = 0 | Centre (2, -3), radius 5 | Halve D and E with signs flipped |
| Circle with diameter endpoints (1, 1) and (7, 9) | x squared + y squared – 8x – 10y + 16 = 0 | Midpoint is the centre |
| Locus equidistant from the origin and x = 4 | y squared = -8(x – 2) | Squaring decides the sign |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Copying the gradient | Writes y – 6 = 2(x – 3) | Perpendicular means the negative reciprocal, so minus one half. |
| Forgetting to flip the centre | Reads the centre as (4, -6) | D equals minus 2h, so h is minus D over 2. |
| Taking the root of F alone | Writes r = root 12 | The radius squared also carries the D and E terms. |
| Guessing the opening direction | Writes y squared = 8(x – 2) | Test one point; the distances expose the sign at once. |
07 Homework (paper, section and question numbers)
- Secondary maths chapter index — The whole locus and circle unit
- Circle properties chapter — The geometric version of the same circle
- Lesson-prep hub — Compare other coordinate geometry chapters
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Locus equidistant from two points | The perpendicular bisector | Wrong means rewrite the distance statement |
| Radius of x squared + y squared – 4x + 6y – 12 = 0 | 5 | Wrong means return to completing the square |
| Locus equidistant from a point and a line | A parabola opening left | Two wrong means repeat act three |
09 Tutor FAQ
Why square the locus equation?
Squaring removes the root and leaves a polynomial; the price is possible extraneous roots, so a numerical check is mandatory.
Can two points determine a circle?
No. The centre only has to lie on the perpendicular bisector, which is a whole line; a third point pins it down.
Why do tangency answers come in pairs?
The distance from the centre uses a modulus, so there is one parallel tangent on each side of the circle.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
