Why the radius meets it at a right angle, argued from shortest distance

What to assume they know

• Counting shared points sorts every line into one of three kinds — a tangent identified by sharing exactly one point with the circle, and what it means for a point to be inside, on, or outside
• A tangent as what a secant becomes when its two crossings merge — the tangent as the limiting position of a secant, and the uniqueness question the activities left open
• That every point of the circle is at distance r from the centre, points inside at less than r, points outside at more
• That a leg of a right triangle is shorter than its hypotenuse
• Pythagoras' theorem, for the numerical work at the end
• Familiarity with proof by contradiction — assume the opposite, derive a clash

What they should be able to do

• State what Theorem 10.1 claims, distinguishing its hypothesis from its conclusion
• Explain why a point of the tangent other than the touching point cannot lie inside the circle, and cannot lie on it
• Convert "lies outside" into a strict inequality between two lengths
• Argue that the radius to the touching point is the shortest segment from the centre to the tangent line
• State the shortest-distance result the proof imports, outline how the appendix proves it by contradiction, and say why contradiction is the appendix's choice rather than the result's requirement
• Deduce the right angle from the minimum, and explain why no measurement is involved
• Explain what Remark 1 settles that the previous topic's activities could not
• Explain what the word normal names, and how it relates to the tangent
• Compute a tangent length from a radius and the distance from the centre, and say which of the two given lengths is the hypotenuse

Where it usually goes wrong

• "You can see the right angle in the figure, so why prove it?" You cannot — Fig. 10.5 carries no right-angle mark, precisely so that nothing is smuggled in. Show the printed page and let the class hunt for the mark that is not there.
• "It is perpendicular because it touches at one point." That is the hypothesis, not the conclusion, and there are four steps between them: no rival point inside, no rival point on the circle, so every rival is farther out, so the radius is the minimum, so it is perpendicular. Skipping the chain is what makes the theorem unmemorable.
• "OQ is longer than OP because the picture shows it slanting." It is longer because Q lies outside the circle and OP is exactly the radius. The slant is a consequence, not a reason.
• "The shortest distance from a point to a line is the perpendicular — that is the definition of distance." It is a theorem, and Appendix 1 proves it by contradiction on p. 237. Distance is defined between two points; extending it to a line is the very thing the theorem licenses.
• "So drawing a perpendicular to any line through the centre gives a tangent." The foot has to land on the circle. The theorem runs from a tangent to a right angle; going the other way is the converse, which is Exercise 10.2 question 5.
• "Normal and tangent are two names for the same line at a point." They are two different lines through the same point, at right angles to each other. The normal runs through the centre; the tangent does not.
• "In a tangent problem you always add the squares." In Exercise 10.1 question 3 the 12 cm reaches the centre and is the hypotenuse, so you subtract. Deciding which length is the hypotenuse is the whole of the arithmetic in this chapter.
• "Theorem 10.1 is about the radius, so it needs the centre to be drawn." It needs the centre to exist, and every circle has one. The commonest exercise form gives you a distance to the centre without drawing the radius at all.

Questions to check understanding

• State Theorem 10.1 and prove it
• Given a radius and the distance from the centre to an external point, find the tangent length, or invert either way
• Multiple choice on the tangent length, with the add-the-squares answer offered as a distractor
• Fill-in-the-blank on how many tangents a circle has, and on the name of the shared point
• Prove the converse — a perpendicular erected on the tangent where it touches must pass through the centre
• Short-answer: explain why the point Q in the proof cannot lie inside the circle
• Prove that the tangents at the two ends of a diameter are parallel, which is Theorem 10.1 applied twice

Examples worth working on the board

Inputs only. Values marked verified are worked out here on the chapter's printed data.

• The wheel that motivates it (§10.2, p. 146, Fig. 10.4). A spoked wheel resting on a horizontal ground line, with the spoke through the contact point drawn and a small square right-angle mark set at the contact. The chapter's own text is careful to say only that the spoke appears square to the road, and then announces the proof. Keep the hedge in view.
• Fig. 10.5, exactly as drawn (§10.2, p. 147). Verified on a close-up taken wide enough to include the caption and the neighbouring text: a circle sits above a horizontal line labelled Y at its left end and X at its right, both ends carrying arrowheads. The circle rests on that line, touching at a point labelled P directly below the centre O. A solid segment runs from O straight down to P. A dashed segment runs from O down and to the right to a point labelled Q, which sits on the line between P and X. There is no right-angle mark anywhere in the figure. That absence is deliberate: the figure must not assert the thing being proved. This is the single most useful visual point in the topic.
• Why Q cannot be inside. The chapter puts the reason in a parenthesis. If Q were inside the circle then the line XY would run from outside, through the interior, and out again — two crossings — so XY would be a secant. But XY was given as a tangent, which shares one point. Contradiction, so Q is not inside.
• Why Q cannot be on the circle. This step is added here; the chapter does not spell it out. If Q were on the circle then XY would share two points with it, P and Q, and again XY would be a secant. So Q is neither inside nor on, and is therefore strictly outside.
• The inequality. Points outside the circle are at distance greater than r from the centre, and OP is exactly r. So OQ > OP, and this holds for every point Q of the line other than P. The chapter states the inequality directly.
• The imported step, and where to find it. Chapter 10 finishes by citing the result that the shortest of the segments from a point to a line is the perpendicular one. The printed citation on p. 147 reads "Theorem A1.7", but in this edition A1.7 is a section number, not a theorem number. Verified by opening the appendix pack: §A1.7 "Proof by Contradiction" begins on p. 234, and the result in question is printed inside it as Theorem A1.2 on p. 237, stated for the segments drawn from a point to the points of a line not through that point. This is a deliberate cross-reference out of this chapter's page range.
• How the appendix proves it (Appendix 1, p. 237, with Fig. A1.5). Take the shortest of the segments and suppose it is not perpendicular. Drop the actual perpendicular from the point onto the line. That perpendicular is now a leg of a right triangle whose hypotenuse is the supposedly shortest segment, so it is shorter — contradicting the choice. Hence the shortest one was perpendicular all along. The appendix lays this out in two columns, statement beside comment. Worth telling the teacher plainly: the result does not need contradiction. Its load-bearing step is the direct comparison that any segment other than the perpendicular is the hypotenuse of a right triangle whose leg is the perpendicular, and is therefore longer — which settles it in one line. The appendix chooses the contradiction form because Theorem A1.2 sits inside §A1.7, whose whole subject is that technique.
• The conclusion. OP is the shortest of the segments from O to points of XY; therefore OP is perpendicular to XY. Nothing has been measured and no picture has been trusted.
• Remark 1 (p. 147). Where the circle passes through the chosen point, the printed remark allows one and only one tangent there. Split that into its two halves, because the theorem supplies only one of them. Uniqueness of the perpendicular from a point to a line rules out a second tangent; it cannot produce a first. The existence half is what the previous topic's Activity 1 supplies, by drawing one.
• Remark 2 (p. 147). The line along which that radius lies, extended past the touching point, is sometimes called the normal there.
• Exercise 10.1 question 3 (p. 147). Inputs: a circle of radius 5 cm, a tangent touching at P, and a point Q on a line through the centre with OQ = 12 cm. Options printed: 12 cm, 13 cm, 8.5 cm, √119 cm. Verified: Theorem 10.1 makes the angle at P a right angle, so OQ is the hypotenuse and PQ² = 144 − 25 = 119, giving PQ = √119 cm, which is about 10.9 cm — the fourth option. The trap is deliberate: 13 is what comes out of 144 + 25 and is the answer to a different question, the one where the 12 is a leg rather than the hypotenuse.
• Exercise 10.1 questions 1 and 2 (iv) (p. 147). How many tangents a circle can have, and the name of the shared point. Verified from this topic: one at each of the infinitely many points of the circle, so infinitely many in total; and the point of contact.
• The converse, flagged as owed (Exercise 10.2 question 5, p. 152 — a deliberate cross-reference two topics ahead). Erect a perpendicular to the tangent where it touches, and show that it has to run through the centre. Verified sketch: if it missed the centre, then that line and the radius would be two different perpendiculars to the tangent at the same point, which is impossible. Theorem 10.1 and this converse are two different statements and students routinely quote one for the other.

Figures to have open

• Fig. 10.5 redrawn (p. 147): circle, centre O, tangent line with X and Y ends, touching point P, rival point Q on the line, the radius solid and the rival segment dashed — and no right-angle mark, until section 10 puts one there. The chapter's own figure; the solid-versus-dashed distinction is load-bearing and must survive the redraw.
• A step-by-step sweep of Q along the tangent line with a live length readout for OQ, so the minimum at P is watched rather than asserted. Standard schematic, not in the book — the chapter draws a single static rival.
• The contradiction picture from Appendix 1 (Fig. A1.5, p. 237): a point above a line, a family of segments down to it, the assumed-shortest one, and the true perpendicular drawn dotted. This lives outside the chapter and should be redrawn as a plain schematic.
• Fig. 10.4's wheel (p. 146), reused from the previous topic so the payoff lands on the same picture that raised the question.
• A right triangle carrying 5 cm, 12 cm and the unknown tangent length, with the right angle at the point of contact. Standard schematic.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 10 "Circles", §10.2, pp. 146–147: the wheel paragraph with Fig. 10.4, the statement of Theorem 10.1, its proof with Fig. 10.5, and Remarks 1 and 2
• Exercise 10.1 questions 1, 2 (iv) and 3, p. 147
• Deliberate cross-reference outside this chapter: the shortest-reach result the proof imports is Theorem A1.2, Appendix 1 "Proofs in Mathematics", p. 237, inside §A1.7 "Proof by Contradiction" which begins on p. 234. The printed citation on p. 147 gives the section number A1.7 in place of the theorem number
• Deliberate cross-reference forward: the converse is set as Exercise 10.2 question 5, p. 152
• The theorem is restated as the second point of the chapter summary, §10.4, p. 153