Deep Dive: The Circle

Provenance (2026-09-29): written against research/veridian/WMA12-verified-facts.md (§2 spec wording, §3 formula-booklet page, §6.3, §7, §8, and the two 2026-09-14 lesson-audit passes); content/veridian/curriculum/wma12/the-circle.ts (read in full, 936 lines, so nothing below re-derives its proofs or repeats its worked chains/marked-solution/MCQs); content/veridian/technique-hub/wma12/forge-exam-technique-guide.md, signal-frequency-and-boundaries.md, codex-formula-and-method-bank.md, and the sibling signal-topic-brief-the-circle.md (all read in full); and content/veridian/question-bank/wma12.ts (grepped — no circle content). Beyond those, this pass independently re-extracted primary Pearson PDFs (pdftotext -layout, cross-checked against -raw on ambiguous lines, with several original micro-examples' arithmetic verified with sympy) for twelve WMA12 sittings not previously checked for this topic anywhere in this course's material: October 2020, October 2021, October 2024, January 2020, January 2022, January 2024, January 2025, June 2022, June 2023, June 2024, and both the main and 01a-regional-variant papers of June 2025 (the existing signal-frequency-and-boundaries.md/forge-exam-technique-guide.md had only the June 2025 main paper's Q2 on record — its 01a sibling paper's own circle question, and the correction that entails, is new to this pass; see §8). Together with the two sittings already on record (January 2021, January 2023), that puts real, directly-verified circle-question coverage at fourteen sittings, fifteen question-instances — see §9. Files read directly: the question paper, mark scheme, and (where separately published) examiner report for each of the twelve newly-checked sittings, held locally under ~/Documents/SOL/Papers/IAL_Paper2_Complete/IAL_Materials/Mathematics/<series>/ and ~/Documents/SOL/Old/P2 Materials/Mathematics/<series>/. This document's job is to be genuinely additive: it does not re-teach the theory the-circle.ts already teaches well, and it spends its length on real exam angles no existing WMA12 lesson or technique-hub document currently names.

#CODEX Deep-Dive: The Circle (WMA12, spec 3.1)

#0. What this document adds, and what it deliberately skips

the-circle.ts already teaches, with full worked derivations: the distance-formula origin of $(x-a)^2+(y-b)^2=r^2$; completing the square to convert general form; the tangent-via-perpendicular-radius method (three real anchors: Jan 2021 Q9, Jan 2023 Q6, plus a chain-drill); the sign-flip, solved-parameter, decimal-not-surd, implicit-differentiation, origin-mislabelled, substitution-direction and area-formula traps. codex-formula-and-method-bank.md already tabulates the three named properties as memorise-and-apply facts. None of that is repeated here.

What follows is new to this course's WMA12 material as of this pass: a real, mark-scheme-confirmed test of the circle's non-degeneracy condition (previously flagged as unverified in codex-formula-and-method-bank.md); a corrected citation for the "unexpected" June 2025 circle question (it is the 01a regional variant's Q9, not the main paper's); the perpendicular-bisector-of-a-chord property used as the primary method for finding an unknown centre, not just as a proved geometric fact (spec property (ii) — flagged by the sibling signal-topic-brief-the-circle.md as having no named anchor anywhere in this course's material until now); the converse of the angle-in-a-semicircle property, used to prove a given segment is a diameter (spec property (i) — the same sibling brief's other flagged gap, now closed); a real trap in which candidates misapply that same property to an angle at the centre rather than the circumference; three genuinely distinct real methods for establishing tangency (perpendicular-gradient, discriminant-of-the-substituted-quadratic, and perpendicular-distance-from-a-point-to-a-line); a two-circles relationship question (do they touch, intersect, or neither) that sits outside spec 3.1's literal three named properties but is real and repeatedly tested; and a synthesis table of every real WMA12 circle question this pass could directly verify — fifteen question-instances across fourteen sittings, up from the three sittings named anywhere in this course's existing material.

Every original numeric micro-example below is labelled VERIDIAN-original and was checked with sympy before being written in; it is not a reproduction of any real question. Every real anchor is described in paraphrase with short attributed quotes under the spec's own copyright discipline — never a reproduced question stem.

#1. The model, in one sentence

A circle is every point at a fixed distance $r$ from a fixed centre; squaring the distance formula gives $(x-a)^2+(y-b)^2=r^2$ directly, and multiplying that out gives general form $x^2+y^2+2gx+2fy+c=0$ with centre $(-g,-f)$ and $r^2=g^2+f^2-c$. The full derivation, and the proofs of all three spec-named properties from one shared right-angled-triangle argument, are in the-circle.ts and are not repeated here.

#2. Building a circle's equation: every real construction route found

Spec 3.1 requires going from a circle's equation to its centre/radius and back. the-circle.ts's teach block already states the centre-plus-point and diameter-endpoints routes; this pass found real anchors for both, plus two routes that are not in that teach block at all.

From a centre and one point on the circumference. The radius is the distance from centre to point. Real anchors: January 2024 Q3(a) (centre $A(3,5)$, passes through $B(8,-7)$, 3 marks) and June 2023 Q3(a) (centre $(2,5)$, passes through $P(8,-3)$, 3 marks for the radius and equation together). June 2023's examiner report records a specific, genuinely interesting error here: some candidates correctly computed the centre-to-$P$ distance (the radius itself) but then treated that number as though it were a diameter and needlessly halved it, scoring nothing — the mirror image of the more familiar diameter-mistaken-for-radius error (§8 below), and confirmation that the diameter/radius distinction, not the arithmetic, is what actually costs marks on this construction route.

From two diameter endpoints. Centre = midpoint of the endpoints; radius = distance from that midpoint to either endpoint. Real anchors: October 2020 Q4 ($P(-11,6)$, $Q(-3,12)$ a diameter, 4 marks for centre/radius + 2 for the equation) and June 2025 Q2 (main paper, 5 marks, already in signal-frequency-and-boundaries.md). October 2020's examiner report records the same family of error as June 2025 Q2's, independently: candidates who used the diameter² directly as the radius² instead of halving first.

From an axis-tangency constraint. Genuinely new to this course's material: WMA12/01a, June 2025 (the regional variant sitting the same day as the main June 2025 paper — see §9 for why this corrects an existing citation), Q9. The real question (paraphrased, not reproduced): a circle has its centre on the $x$-axis, touches the $y$-axis, and passes through a given point; sketch it (1 mark), then find its equation (4 marks). The examiner report calls this "another unexpected question that candidates did not seem fully prepared for" and states the key step is realising the radius equals the centre's own $x$-coordinate (since the centre sits on the $x$-axis and the circle touches the $y$-axis, the perpendicular distance from centre to the $y$-axis — which equals the radius, by the radius-⊥-tangent property applied to the coordinate axis itself — is just $|a|$ for a centre $(a,0)$). The real mark scheme confirms three independently valid routes credited in full: (1) set up $(x\mp a)^2+y^2=a^2$ symbolically and substitute the given point to solve for $a$; (2) draw the right-angled triangle formed by the centre, the given point, and the foot of the perpendicular from the centre to the vertical through that point, and apply Pythagoras directly; (3) use the perpendicular-bisector-of-a-chord method (§4.2 below).

VERIDIAN-original micro-example, same structure, different numbers: a circle has centre on the $x$-axis, touches the $y$-axis, and passes through $(2,4)$. Centre $(a,0)$, radius $a$ (since it touches the $y$-axis): $(2-a)^2+4^2=a^2 \Rightarrow 4-4a+a^2+16=a^2 \Rightarrow 20=4a \Rightarrow a=5$. Equation: $(x-5)^2+y^2=25$. (Checked with sympy.)

The non-degeneracy condition — now confirmed real-exam-tested. codex-formula-and-method-bank.md flags $g^2+f^2-c>0$ as "a genuine mathematical fact, not a confirmed exam-tested one." This pass found a direct counter-example: October 2021, Q6(i)(b), 2 marks. Given $C_1$: $x^2+y^2+10x-12y=k$, part (a) finds the centre $(-5,6)$ (2 marks); part (b) asks for "the possible range in values for $k$" directly — the real mark scheme's method line is $k+(\pm5)^2+(\pm6)^2>0$, answer $k>-61$. This single question resolves the previously-flagged gap: the boundary case is real, tested, and worth exactly 2 marks on its own, immediately after the centre step, on a question that also (in part (ii)) builds a second circle from diameter endpoints and substitutes a point with one unknown coordinate into it to solve for that coordinate — a genuine combination of two of this section's construction routes in one question.

VERIDIAN-original micro-example: for $x^2+y^2-6x+8y=k$ to be a genuine circle, complete the square: $(x-3)^2+(y+4)^2=k+25$, so $k+25>0$, i.e. $k>-25$. (Checked with sympy.)

#3. The perpendicular bisector of a chord — as a method, not just a proof

the-circle.ts proves this property (spec item (ii)) and tests reading it in one MCQ, but has no worked example using it to solve a multi-mark problem. This pass found three real anchors that do.

As the primary method for finding an unknown centre. January 2025, Q6, 8 marks (4+1+3) — the cleanest real anchor found this pass. Two points $A(-1,2)$ and $B(11,6)$ lie on a circle with centre $P$; $M$ is the midpoint of $AB$; the line through $M$ and $P$ is asked for directly (4 marks — this line is the perpendicular bisector of chord $AB$, by the spec property, and the question does not need to say so explicitly for the property to be the entire method). Given $P=(7,k)$, find $k$ (1 mark, substitute $x=7$ into the line just found) and then the circle's equation (3 marks, radius from $P$ to $A$ or $B$). The examiner report records a real, sharp trap distinct from anything in the-circle.ts's existing taxonomy: some candidates "wrongly assum[ed] that $ABP$ was a right-angled triangle" — misapplying the angle-in-a-semicircle property to an angle at the centre ($P$), when that property only ever concerns an angle at a point on the circumference subtended by a diameter. A second confirmed error on the same question: using $AB^2$ directly as the radius², treating a chord's length as though it were a diameter or radius outright.

As an alternative construction route. The June 2025 01a axis-tangency question (§2 above) explicitly credits a third method built on exactly this property: since the given point and the circle's tangency point (on the $y$-axis) are both known points on the circle, the perpendicular bisector of the chord joining them must pass through the centre — and since the centre is also known to lie on the $x$-axis, intersecting that bisector with the $x$-axis finds the centre directly, with no algebra involving an unresolved constant at all.

VERIDIAN-original micro-example demonstrating this exact route on the §2 axis-tangency circle above: it touches the $y$-axis at the origin $(0,0)$ (the foot of the perpendicular from centre $(5,0)$) and passes through $(2,4)$ — both real points on the circle. Midpoint of that chord: $(1,2)$. Chord gradient $\frac{4-0}{2-0}=2$, so the perpendicular bisector has gradient $-\tfrac12$: $y-2=-\tfrac12(x-1)$. Setting $y=0$: $x=5$ — the same centre found in §2 by the direct substitution method. (Checked with sympy.)

Finding a chord's own equation given its length and position. Two independent real anchors, both testing the converse direction — using the chord-bisection property (via Pythagoras: half-chord² + centre-to-chord-distance² = radius²) to locate a chord rather than to locate a centre: January 2020, Q6(c) (3 marks — a horizontal chord $y=p$, $p$ negative, of length 4, on a circle already found in part (a)) and January 2024, Q3(b) (3 marks — a horizontal chord $MN$ above the $x$-axis, length $4\sqrt{22}$, on a circle found in part (a)). Both real mark schemes credit multiple routes: the Pythagoras half-chord method; direct substitution of the chord's $x$-endpoints (found from $x=a\pm\sqrt{r^2-(\text{distance})^2}$) back into the circle's equation; or, on January 2020's version specifically, using "the difference between the roots is 4" applied to the quadratic obtained by substituting $y=p$ into the circle equation directly. January 2024's mark scheme additionally credits a cosine-rule route through the triangle formed by the chord's midpoint, one endpoint, and the centre.

VERIDIAN-original micro-example: circle $(x-2)^2+(y-5)^2=25$; find the equation of the horizontal chord of length $8$ lying below the centre. Half-chord $=4$; perpendicular distance from centre to chord $=\sqrt{25-16}=3$; chord at $y=5-3=2$. Check: at $y=2$, $(x-2)^2+9=25\Rightarrow x=-2$ or $6$, chord length $=8$. ✓ (Checked with sympy.)

#4. Radius ⊥ tangent — three genuinely different real methods

the-circle.ts teaches the perpendicular-gradient method in full and names implicit differentiation as a real but weaker alternative (very few real candidates who tried it made progress — Jan 2021 Q9). This pass found two further real methods, both suited specifically to the case the gradient method cannot use directly: when the tangent's own point of contact on the circle is not yet known, only a fixed external point the tangent passes through.

Discriminant of the substituted line = 0. Real anchors: January 2020, Q6(b) (2 marks — a horizontal tangent $y=k$; real alternative credited: substitute $y=k$ into the circle equation and set the discriminant to zero); June 2022, Q10 (12 marks across three parts — circle centre $X(3,5)$ with a symbolic radius $r$; line $y=2x+k$; part (a), 3 marks, is a 'show that' proving the resulting quadratic $5x^2+(4k-26)x+k^2-10k+34-r^2=0$; part (b), 3 marks, uses the discriminant-zero tangency condition to derive $5r^2=(k+p)^2$ for a constant $p$ to be found; part (c), 6 marks, adds a further geometric constraint — the tangent's $y$-intercept $A$ and point of contact $B$ satisfy $AB=2r$ — solved via Pythagoras in the right-angled triangle formed by the centre, $A$, and $B$); and October 2024, Q10(b) (3 marks — a tangent family $y=mx+1$ through a fixed point, substituted into a genuine general-form circle to reach a printed quadratic-in-$m$, $2m^2-7m-22=0$, as a 'show that', then part (c), 4 marks, solves for both values of $m$ and both points of tangency). This last anchor's real mark scheme also names a fourth, entirely separate method for the same step: setting the perpendicular distance from the centre to the line (via $d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$) equal to the radius directly — avoiding a quadratic-in-$x$ altogether.

VERIDIAN-original micro-example (deliberately simple, to isolate the method): circle $x^2+y^2=25$ (centre origin, $r=5$); tangent lines from the external point $(13,0)$, family $y=m(x-13)$. Substituting: $x^2+m^2(x-13)^2-25=0 \Rightarrow (1+m^2)x^2-26m^2x+(169m^2-25)=0$. Discriminant $=0$: $(26m^2)^2-4(1+m^2)(169m^2-25)=100-576m^2=0 \Rightarrow m=\pm\tfrac{5}{12}$. (Checked with sympy; the $5,12,13$ Pythagorean triple falling out — tangent length $\sqrt{13^2-5^2}=12$ — is an independent arithmetic check that the result is genuinely consistent, not a coincidence built into the question.)

Two circles, one line, one discriminant sign. §7 below shows the same discriminant idea extended to testing whether two circles meet at all — with the sign of the discriminant, not just whether it is zero, doing the work.

#5. The angle in a semicircle — the converse, and a real, sharp trap

the-circle.ts proves the forward direction (any triangle on a diameter has a right angle at the third vertex) and tests reading it in two MCQs, but — as signal-topic-brief-the-circle.md explicitly flags — no existing WMA12 material in this course names a real anchor using it as a proof method. This pass found two.

Proving a given segment is a diameter, from three given points. June 2022, Q3(ii), 5 marks: given three points $A$, $B$, $C$ on a circle, prove $AB$ is a diameter. The real mark scheme credits three independently valid routes: (Way 1) show the gradients of $AC$ and $BC$ multiply to $-1$, hence the angle at $C$ is $90°$, hence — by the converse of the semicircle property — $AB$ is a diameter; (Way 2) the same conclusion via Pythagoras on the three side lengths instead of gradients; (Way 3) find the midpoint $M$ of $AB$ directly and show $MA=MC$ (both automatically equal $MB$), i.e. show $M$ is equidistant from all three points and is therefore the circle's own centre — which, since $M$ is also $AB$'s midpoint, makes $AB$ a diameter without ever computing an angle at all.

Using it to construct the circle itself. January 2022, Q6, 8 marks (2+3+3): given $P$, $Q$, $R$ on a circle, part (a) asks to show $\angle PQR=90°$ (2 marks, same three method families as above); part (b) uses that result to state that $PR$ — not $PQ$, since the right angle sits at $Q$ — is a diameter, giving the centre (its midpoint, B1) and radius (M1A1) directly, with no completing-the-square step anywhere in the question. Part (c) then asks for the tangent at the point $S$ on the circle for which the distance $QS$ is greatest — real geometric fact: the farthest any two points on a circle can be from each other is the diameter, so $S$ must be diametrically opposite $Q$, i.e. the centre is the midpoint of $Q$ and $S$. The real mark scheme's own method note describes this as a vector-doubling shortcut: $\vec{QS}=2\vec{QC}$ (finding $S$ by moving from $Q$ through the centre and out the same distance again, rather than solving simultaneous equations).

VERIDIAN-original micro-example for both ideas together: $A(0,0)$, $B(6,8)$, $C(6,0)$. $\vec{CA}=(-6,0)$, $\vec{CB}=(0,8)$; dot product $=0$, so the angle at $C$ is $90°$. Midpoint of $AB$: $M(3,4)$. $MA=\sqrt{9+16}=5$, $MC=\sqrt{9+16}=5$ — equal, confirming $M$ is the centre and $AB$ (length $10=2\times5$) is a diameter. Now take a different point on the same circle, $Q=(8,4)$ (check: $(8-3)^2+(4-4)^2=25$ ✓); the point diametrically opposite $Q$ is $S=2M-Q=(6,8)-(8,4)=(-2,4)$ (check: $(-2-3)^2+(4-4)^2=25$ ✓). (Both checked with sympy.)

The real trap this property invites. January 2025 Q6's "$ABP$ right-angled" error (§3 above) is the natural failure mode of learning the semicircle property only in its forward direction: a triangle formed by two points on a circle and its centre is not covered by the property at all (it is generally isosceles, not right-angled, since two of its sides are radii) — the property strictly requires the right angle's vertex to be a third point on the circumference, with the other two points forming a diameter through the centre, not the centre itself as one of the three vertices.

#6. Beyond spec 3.1's three named properties: two circles

Spec 3.1 names exactly three circle properties, all concerning a single circle. This pass found a real, repeatedly-tested question type that sits outside that literal list: whether two given circles touch, intersect, or do neither — flagged here explicitly as an extension beyond the spec's own named content, not asserted as one of its three bullets, but real and worth knowing about.

June 2024, Q7, 6 marks (3+3): $C_1$ in general form (3 marks, centre/radius as usual), $C_2$ given directly in centre-radius form; part (b), under a "you must show detailed reasoning" instruction, asks to prove the two circles neither touch nor intersect. The real mark scheme credits two routes: (1) compare the distance between the two centres against the sum of the two radii, with both values computed to a decimal and the inequality stated explicitly — a bare inequality with no supporting numbers does not earn the mark; (2) subtract the two circle equations to eliminate both quadratic terms, producing a straight line (the radical axis), substitute that line into either circle equation to reach a three-term quadratic, and show its discriminant is negative — i.e. the same discriminant tool §4 uses for tangency ($=0$), now used with the opposite sign to prove no intersection at all exists.

October 2021, Q6(ii), 4 marks, is a related but distinct real anchor: a second circle $C_2$ is built from two given diameter endpoints, and a third point with one unknown coordinate is given as also lying on $C_2$ — testing the diameter-endpoints construction (§2) rather than the touch/intersect relationship, but confirming that "two circles in one question" is itself a recurring real shape, not a one-off.

VERIDIAN-original micro-example: $C_1$: $x^2+y^2-4x-6y+4=0$ (centre $(2,3)$, $r_1=3$); $C_2$: $(x-10)^2+(y-15)^2=64$ (centre $(10,15)$, $r_2=8$). Distance between centres $=\sqrt{8^2+12^2}=\sqrt{208}=4\sqrt{13}\approx14.42$; sum of radii $=11$. Since $14.42>11$, the circles are separate — neither touching nor intersecting. (Checked with sympy.)

#7. The diameter/radius confusion — a single error family, two opposite directions

forge-exam-technique-guide.md already names this trap from June 2025 Q2 alone and flags it as absent from the-circle.ts's own taxonomy. This pass found the same underlying confusion recurring in both possible directions, across four sittings:

• Treating a diameter's length as though it were already the radius (forgetting to halve): June 2025 Q2 (main paper) and October 2020 Q4 — both diameter-endpoints constructions (§2), both confirmed independently.
• Treating an already-correct radius as though it were a diameter (halving when there was nothing to halve): June 2023 Q3(a) — a centre-plus-point construction (§2), where the centre-to-point distance computed is the radius outright, and some candidates halved it anyway, scoring nothing.
• Treating a chord's length as though it were the radius directly, with no distance-to-radius relationship applied at all: January 2025 Q6(c) — using $AB^2$ (a chord, not even a diameter) as the radius².

The common thread across all four: none of these is an arithmetic slip. Each is a genuine uncertainty about which real-world length a given number represents inside a circle construction — diameter, radius, or an unrelated chord — and the fix in every real mark scheme is the same: identify what geometric object each given length actually is before choosing whether to halve, square, or substitute it.

#8. Corrections this pass makes to this course's own existing material

Per this course's standing discipline, corrections are stated plainly rather than silently folded in.

The June 2025 "unexpected" circle question is the 01a regional variant's Q9, not the main paper's. WMA12-verified-facts.md §6.3 cites "June 2025, Q9" for the axis-tangency question without specifying which paper. Re-extracting the real main-paper question paper directly (wma12-01-que-20250514.pdf) shows its own Q9 is an unrelated 3-D open-container optimisation question (surface area, differentiation) — already correctly identified as such, independently, in forge-exam-technique-guide.md. The axis-tangency circle question is confirmed, by directly reading wma12-01a-que-20250514.pdf and its mark scheme/examiner report, to be Q9 of the 01a regional variant paper sat the same day. The facts bank's own citation is not fabricated — the quote is real and correctly transcribed — but it omits the paper variant, and one existing technique-hub sibling (signal-topic-brief-the-circle.md) inherited the ambiguity without resolving it. §2 and §3 above cite this anchor correctly as "WMA12/01a (June 2025 regional variant), Q9."

The circle's non-degeneracy condition is confirmed real-exam-tested. codex-formula-and-method-bank.md flags $g^2+f^2-c>0$ as untested in the material reviewed at the time. §2 above names the real counter-example directly: October 2021 Q6(i)(b), 2 marks, "state the possible range in values for $k$."

#9. Synthesis: every real WMA12 circle question this pass could verify directly

Confirmed directly against a primary Pearson PDF (question paper and/or mark scheme, cross-checked against the examiner report where one exists) in this pass or an earlier one this course has recorded — fifteen question-instances across fourteen sittings (June 2022 is the one sitting carrying two separate circle questions, Q3(ii) and Q10). Never presented as a percentage or a completed sweep of the archive's full 20-series scope.

Not deep-dived here despite being real and on-topic (checked, judged not to add a new angle beyond what §§1–8 already cover): October 2022 (a routine tangent-from-a-point question) and October 2023 Q7 ("Circle rules and formulae" — the radius-⊥-tangent property set as a proof/'show detailed reasoning' question, reinforcing the same 'ag' discipline the-circle.ts already teaches from Jan 2021 Q9, not a new sub-case).

#10. What remains genuinely unverified

• June 2019 and October 2019 were not checked directly for circle content in this pass (this pass's own new-checking scope was the twelve sittings listed in the provenance line); WMA12-verified-facts.md §8 already flags both as under-read for other reasons.
• The exact wording of the June 2024 Q7 "detailed reasoning" instruction's own boilerplate (whether it is the same bold per-question warning forge-exam-technique-guide.md documents elsewhere) was read directly on the question paper but not independently cross-checked against that document's own transcription conventions.
• Whether the two-circles question type (§6) recurs beyond the two sittings named here (June 2024, October 2021) has not been checked against the remaining sittings in the fourteen this pass covered that turned out not to feature it, nor against the sittings this pass did not check at all — "seen in 2 of 14 sittings checked for circle content this pass," not a claim about the full 20-series archive.
• This document's own arithmetic for every VERIDIAN-original micro-example was checked with sympy in this session; the real past-paper numbers quoted (mark totals, coordinates, the printed 'show that' targets) are transcribed from the primary PDFs directly, not recomputed independently from scratch beyond the spot checks shown in §§2–7.