Fixed distance from a fixed point, turned into an equation

What to assume they know

• The distance formula for two points in the plane, which §9.1 of Chapter 9 (p. 151) recalls and this section uses without re-deriving
• Expanding a squared binomial, and recognising an expansion when it is only partly present
• One surface, four curves, chosen by the angle of the cut — that the circle is the section produced when the cutting plane is perpendicular to the cone's axis, which is the sense in which it belongs in this chapter at all
• Solving a small simultaneous system in three unknowns
• Substituting a point into an equation to test membership

What they should be able to do

• State Definition 1 and name its two ingredients
• Derive a circle's equation from its centre and radius, starting from the distance formula, and say where the squaring step is used
• Write down the equation given centre and radius, with correct signs
• Recover centre and radius from an expanded second-degree equation by completing the square
• Explain why completing the square is guaranteed to reproduce the derivation in reverse
• Count the unknowns in the general equation and match them against the number of conditions a problem supplies
• Set up and solve a three-condition problem for an unknown circle
• Decide whether a given point lies inside, outside or on a given circle
• Recognise when the recovered right-hand side fails to be a positive number, and say what the equation then describes

Where it usually goes wrong

• "Completing the square is a trick you memorise." It is the expansion of (x − h)² run backwards, and the derivation on p. 180 is where that expansion came from in the first place. Show the forward expansion and the backward completion on the same screen and the "trick" disappears.
• "Centre (−3, 2) gives (x − 3)² + (y + 2)²." The commonest error in the section. The template subtracts the centre's coordinates, so a negative coordinate produces a plus sign. Example 2 exists to make exactly this happen.
• "Any equation with x² and y² in it is a circle." Two conditions are needed: the x² and y² coefficients must match, and what is left after completing the squares must be positive. Item 9 of Exercise 10.1 fails the first condition until you divide through; item 5's radius can fail the second. Note that §10.3, pp. 179–181, prints no statement of either condition — checked on the page images — so a student who has only read the section has not been told.
• "Two points determine a circle." Three numbers are unknown, so three independent conditions are needed. Example 4 supplies two points and a line for the centre; every item from 10 to 14 is the same accounting with different ingredients. Getting students to count before they compute is the point.
• "r² is one of the things you solve for first." In Example 4 it is the last thing found, because subtracting the two point-conditions eliminates it. Set up the elimination and r² takes care of itself.
• "A radius can be any expression you like." Item 5's √(a² − b²) is real only under a condition on a and b, and it can be zero. An equation of circle form is not automatically a circle.
• "Testing a point needs a diagram." It needs one substitution and a comparison. Item 15 is the whole method.

Questions to check understanding

• Write the equation given centre and radius, including negative and fractional centres
• Find centre and radius from an expanded equation, including one whose leading coefficients are not 1
• Find a circle from three conditions, at least one of which is a line constraining the centre
• Decide whether a point is inside, outside or on a given circle
• Find a circle through the origin with given axis intercepts
• Short-answer: state what must be true of the constant term for an equation in circle form to describe an actual circle

Examples worth working on the board

Values marked verified are worked out here from the chapter's stated data; nothing here was taken from an answer key.

• Fig 10.11 (p. 180). A circle with centre marked O, three points marked on it, the three radii drawn, and the equality of the three lengths written under the figure. The lettering is inside the artwork. This is Definition 1 as a picture and nothing more.
• Fig 10.12 (p. 180). The same circle placed on coordinate axes: centre C at (h, k), a general point P at (x, y), origin O marked at the corner. The step from Fig 10.11 to Fig 10.12 is the whole method of the chapter — put the figure on axes, then let the distance formula speak.
• The derivation (p. 180). Centre (h, k), radius r, general point (x, y). The definition asserts the length CP is r; the distance formula turns that into a square root set equal to r; squaring clears it. Verified as reversible: both sides are non-negative before squaring, so no solution is gained or lost. This is the reason the explanation can say squaring costs nothing here, and it is not true of every squaring step in this chapter — the ellipse and hyperbola derivations square twice and need a sign check.
• Example 1 (p. 180). Centre (0, 0), radius r. Inputs only.
• Example 2 (p. 180). Centre (−3, 2), radius 4. Verified: (x + 3)² + (y − 2)² = 16. The sign reversal is the teachable moment — a centre coordinate of −3 produces a plus sign in the bracket.
• Example 3 (p. 180). The equation x² + y² + 8x + 10y − 8 = 0. Verified by an added completion of the squares: the x-group needs 16 and the y-group needs 25, and 8 + 16 + 25 = 49, so the circle has centre (−4, −5) and radius 7.
• Example 4 (p. 181). A circle through (2, −2) and (3, 4), centre on the line x + y = 2. Verified by an added elimination: subtracting the two point-conditions removes r² and h² and k² together and leaves 2h + 12k = 17; with h + k = 2 this gives k = 1.3 and h = 0.7, and substituting back gives r² = 1.69 + 10.89 = 12.58. These agree with the values printed on p. 181. The structural point is that r² never had to be found until the end — it is the dependent unknown.
• Exercise 10.1 items 1–5 (p. 181), centre and radius given, equation wanted: (0, 2) and 2; (−2, 3) and 4; (½, ¼) and 1/12; (1, 1) and √2; (−a, −b) and √(a² − b²). Verified, item 5 is the interesting one: the radius is an expression, not a number. It is real only when a² ≥ b², and it is zero when |a| = |b|, at which point the equation describes the single point (−a, −b) — the degenerate case from Cuts through the vertex, where the curve degenerates arriving in the algebra. The chapter poses the item without comment.
• Exercise 10.1 items 6–9 (p. 181), centre and radius wanted. Each equation with the reading I derived from it:
  • item 6, (x + 5)² + (y − 3)² = 36 — already completed, so centre (−5, 3) and radius 6 can be read straight off. It is in the set to show the target form.
  • item 7, x² + y² − 4x − 8y − 45 = 0 — verified: centre (2, 4), radius √65.
  • item 8, x² + y² − 8x + 10y − 12 = 0 — verified: centre (4, −5), radius √53.
  • item 9, 2x² + 2y² − x = 0 — verified: divide through by 2 first, because the derivation on p. 180 assumes the two squared terms carry coefficient 1. After that the centre is (¼, 0) and the radius ¼.
• Exercise 10.1 items 10–14 (p. 181), the three-condition family: through (4, 1) and (6, 5) with centre on 4x + y = 16; through (2, 3) and (−1, 1) with centre on x − 3y − 11 = 0; radius 5, centre on the x-axis, through (2, 3); through the origin, with axis intercepts a and b; centre (2, 2) through (4, 5). Verified, item 12 admits two answers: a centre (h, 0) with (2 − h)² + 9 = 25 gives (2 − h)² = 16, so h = 6 or h = −2, and there are two circles of radius 5 meeting the conditions. The question asks for "the equation" in the singular. Verified, item 13: the circle passes through (0, 0), (a, 0) and (0, b), and those three points give x² + y² − ax − by = 0 directly.
• Exercise 10.1 item 15 (p. 181). Does (−2.5, 3.5) lie inside, outside or on x² + y² = 25? Verified: 6.25 + 12.25 = 18.5, which is below 25, so the point is inside. This is the only item in the exercise that uses the equation as a test rather than as an object to be found.

Figures to have open

• Fig 10.11 (p. 180) redrawn: one centre, three marked points, three radii, the equality stated. Standard schematic.
• Fig 10.12 (p. 180) redrawn: the circle on axes with C at (h, k) and P at (x, y), and the right triangle with legs x − h and y − k drawn in, which the printed figure does not show. That triangle is the distance formula made visible and is what turns section 3 from an assertion into a picture.
• A single identity panel showing (x − h)² expanded to x² − 2hx + h², with arrows running both ways, for section 8.
• A number line of the recovered constant, marked positive / zero / negative, with a circle, a point and nothing drawn beneath the three regions.
• No printed figure exists for the two circles in Exercise 10.1 item 12; drawing both on one pair of axes is the cheapest way to show why the singular "the equation" is misleading.

Where this sits in the book

• NCERT Mathematics, Textbook for Class XI, Chapter 10 "Conic Sections", §10.3 Circle (pp. 179–181): Definition 1 (p. 179), the derivation and Examples 1–3 (p. 180), Example 4 and Exercise 10.1 (p. 181)
• Figures: Fig 10.11 and Fig 10.12, both on p. 180
• Deliberate backward reference outside the chapter: the distance formula is used without proof and belongs to Chapter 9, §9.1, p. 151
• Deliberate reference inside the chapter: the circle's place among the conic sections is fixed at §10.2.1, p. 177
• The chapter's Summary (p. 205) restates Definition 1 and the centre-and-radius equation, and nothing else about circles — no eccentricity, no latus rectum, no reduction of a general second-degree equation