Three points not in a line: exactly one circle (Theorem 1)

What to assume they know

• Two points: infinitely many circles, centres on the perpendicular bisector — the centres of all circles through two points fill the perpendicular bisector of the segment joining them, in both directions
• Turning an observation into a definition: the circle as a locus — circle, centre, radius, chord
• That two lines in a plane either meet at exactly one point or are parallel
• What collinear means, and how to test three drawn points for it
• Classifying a triangle as acute-, obtuse- or right-angled from its angles
• Ruler-and-compasses construction of a triangle from given sides and angles, and of a perpendicular bisector

What they should be able to do

• Explain why three points lying on one straight line cannot all be on one circle
• Show that for three collinear points the perpendicular bisectors of two of the segments are parallel, and say what that rules out
• State Theorem 1, and identify which words of the statement carry the existence claim and which the uniqueness claim
• Locate the centre of the circle through three given non-collinear points by intersecting two perpendicular bisectors, and say why the third bisector adds nothing
• Explain why exactly one intersection point gives both existence and uniqueness
• Use the terms circumcentre, circumcircle, circumscribe and inscribed correctly, and say which figure each describes
• Predict from a triangle's angles whether its circumcentre falls inside, outside, or on the triangle, and check the prediction by construction
• State where the circumcentre sits for a right-angled triangle, and connect it to the right-angle result that arrives later in the chapter
• Say whether two different triangles can share one circumcircle

Where it usually goes wrong

• "Any three points lie on some circle." Not if they are collinear. The chapter asks the reader to explain the failure before it states the theorem, and an explanation that states the theorem first has thrown away the pedagogy.
• "Collinear points fail because the circle would have to be infinitely big." That is a picture, not a reason. The reason is that the two perpendicular bisectors are parallel, so there is no point that is equidistant from all three.
• "Three bisectors, so three conditions." Two suffice. If O is equidistant from A and B and equidistant from A and C, it is automatically equidistant from B and C. The chapter uses exactly two and lets the figures show the third.
• "The circumcentre is inside the triangle." True only for acute-angled triangles. Figs. 5.6 and 5.7 exist precisely to break this, and the two construction questions in Exercise Set 5.1 are designed so that one lands inside and one outside.
• "Uniqueness needs a separate argument." It does not, here. The bisectors meet at one point only, so there is one candidate centre; the radius is then forced. Both halves come out of the same intersection.
• "The circumcentre is the centre of the triangle." The triangle has several centres serving different purposes; this one is defined by being equidistant from the vertices, and it can sit outside the triangle entirely — which no student's mental image of a "centre" allows.
• "One circumcircle, one triangle." No: rotate the triangle inside its circle and you get another triangle congruent to it on the same circle. Think, Draw and Infer Q2 is asking for exactly this.

Questions to check understanding

• Construct the circumcircle of a triangle given two angles and the included side, or given three sides, and state whether the centre is inside or outside
• Given a triangle's angles, predict where the circumcentre lies, with a reason
• Prove that exactly one circle passes through three non-collinear points
• Explain why no circle passes through three collinear points
• Show that the perpendicular bisectors of the sides of a triangle all pass through one point
• State where the circumcentre of a right-angled triangle lies, and find the circumradius from the legs
• One-mark: the greatest number of points in which a straight line can cut a circle
• Given a circle, produce a second triangle congruent to a given inscribed one and inscribed in the same circle

Examples worth working on the board

The chapter prints no answers, so every value below marked verified is worked out here on the chapter's own stated inputs.

• Fig. 5.5 (p. 96, caption names the circumcircle of triangle ABC). Triangle ABC inscribed in a circle with all three perpendicular bisectors drawn as long lines running out past the circle, each labelled in type beside the figure — "Perpendicular bisector of AB", "of BC", "of AC". Right-angle ticks are marked where the bisectors meet the sides. The centre O is inside the triangle. This is an acute-angled case.
• Fig. 5.6 (p. 97). The same construction on an obtuse triangle: O is drawn clearly outside the triangle, and the caption says so. The three bisector labels are printed again.
• Fig. 5.7 (p. 97). The right-angled case: O sits on the hypotenuse, at its midpoint, and the caption states that. Note as a check: this figure is the chapter quietly pre-announcing the Corollary on p. 110 — the right-angle result and this circumcentre position are the same fact seen from two sides. Say so.
• Exercise Set 5.1, Q1 (p. 98). Inputs: AB = 5 cm, ∠A = 70°, ∠B = 60°; draw the circumcircle and say whether the centre is inside or outside. Verified: ∠C = 50°, so all three angles are acute and the centre falls inside.
• Exercise Set 5.1, Q2 (p. 98). Inputs: AB = 5 cm, ∠A = 100°, AC = 4 cm; same question. Verified: the angle at A exceeds 90°, so the triangle is obtuse-angled and the centre falls outside.
• Exercise Set 5.1, Q3 (p. 98). Inputs: AB = 6 cm, BC = 7 cm, CA = 7 cm; draw the circumcircle, call the centre O, measure OA, OB, OC. The measurement is the point: they must come out equal, because all three are radii. Verified: the common value is about 3.87 cm.
• Exercise Set 5.1, Q4 (p. 98). The least possible radius through two points — half the distance between them. Belongs mainly to Two points: infinitely many circles, centres on the perpendicular bisector; useful here as the two-point case the three-point case is being compared against.
• Think, Draw and Infer, Q1 (p. 98). Inputs: three collinear points A, B, C. The chapter asks for a point P with PA = PB = PC, asks what the bisectors of AB and BC do, asks whether a circle can pass through collinear points, and asks whether a line can cut a circle in three distinct points. Verified: no such P exists; the two bisectors are parallel; no circle passes through three collinear points; a line meets a circle in at most two points — which is the same fact restated, and worth making the explanation's closing beat of section 3.
• Think, Draw and Infer, Q2 (p. 98). Can other triangles congruent to ΔABC share its circumcircle? Verified: yes, and infinitely many — turn ABC about the centre by any angle and the three vertices stay on the circle. This is the rotational symmetry of Total rotational symmetry, and why every diameter is an axis of reflection being cashed in, and it is worth naming as such.

Figures to have open

• Figs. 5.5, 5.6 and 5.7 redrawn as a three-panel comparison on one row, each with its bisectors labelled and its centre marked. These are the chapter's own figures (pp. 96–97) and the comparison across them is the argument of section 10; drawn separately and pages apart, as the book has them, the pattern is easy to miss.
• A collinear-failure figure: three points in a line, two perpendicular bisectors, visibly parallel. Standard schematic; the chapter asks for this at Think, Draw and Infer Q1 and does not draw it.
• A construction movement for section 5 and section 7: two bisectors appearing, crossing, and the compasses opening to the first vertex. Standard schematic.
• A rotation figure for section 11's last beat: one circle, one triangle, the triangle turned to a second position with its vertices still on the circle. Standard schematic; the chapter poses the question and draws nothing.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 5, "I’m Up and Down, and Round and Round", §5.3, printed heading "How Many Circles?". The three-point material runs from the top of p. 96 through p. 97; Theorem 1 and its argument are on p. 96, and the circumcircle vocabulary and the three position cases are on p. 97.
• Fig. 5.5 with its caption, p. 96. Figs. 5.6 and 5.7 with their captions, p. 97.
• Exercise Set 5.1, Q1–Q4, p. 98. Think, Draw and Infer, Q1–Q2, p. 98 — the printed block heading is not numbered and can be named but not cited by number.
• Forward pointer inside the chapter: Fig. 5.7's right-angled case is the same fact as the Corollary on p. 110; see The Corollary: a diameter stands on a right angle wherever you take the point.
• Chapter Summary, p. 117 — the bullet on the unique circle through three points not in a line.