Why the two from an outside point are always equally long

What to assume they know

• Why the radius meets it at a right angle, argued from shortest distance — Theorem 10.1, the right angle between a tangent and the radius at the point of contact
• None, one or two, depending on where the point sits — that an external point carries exactly two tangents, and what the length of a tangent means
• The RHS congruence criterion for right triangles, and CPCT
• Pythagoras' theorem
• That the base angles of an isosceles triangle are equal, and that the angles of a triangle total 180°
• The Class IX result that the perpendicular dropped from the centre onto a chord bisects it
• AA similarity, for the second route through Example 3

What they should be able to do

• State what Theorem 10.2 claims and identify the two triangles its proof compares
• Explain why the right angles are available, naming the theorem that supplies them
• Apply RHS congruence and read off the equality of the tangent lengths by CPCT
• Reproduce the Pythagoras route to the same conclusion and say what it makes visible that the congruence route hides
• Deduce that the segment to the centre bisects the angle between the two tangents
• Use the equality to prove that a chord of the larger of two concentric circles, touching the smaller, is cut in half at the touching point
• Relate the angle at the external point to the angle subtended at the centre by the segment joining the two touching points
• Find a tangent length from a chord and a radius by two independent routes and check that they agree
• Use equal tangent lengths additively on a quadrilateral or a triangle drawn round a circle

Where it usually goes wrong

• "The two tangents are equal because the picture is symmetric." The symmetry is the conclusion, not a premise. Nothing in the setup says the figure is symmetric until the congruence has been established. Do not claim the artwork is complicit in this: read, Fig. 10.7 is drawn mirror-symmetric about PO, with Q above and R below in matching positions and PQ and PR plainly equal to the eye. The picture gives the game away; the argument still has to be made.
• "Two sides and an angle always give congruence." They do not; the non-included case is genuinely ambiguous. RHS works because the angle is right. If the explanation lets this slide, the theorem's proof becomes a ritual.
• "PQ = PR makes triangle PQR equilateral." It makes it isosceles. Example 2 depends on it being isosceles and on nothing more.
• "The angle bisector result is a second theorem to learn." It is the same congruence read on a different pair of corresponding parts. Learning it separately doubles the memory load and hides where it comes from.
• "The angle at the external point plus the angle at the centre make 90°." They make 180°. Question 2's data — 110° at the centre giving 70° at the point — is the check; 90° is what the two halves make in question 9's configuration, and mixing them up is the commonest slip in this exercise.
• "AB + CD = AD + BC holds for any quadrilateral." Only when a circle can be drawn touching all four sides. The hypothesis is doing real work, and question 11 is exactly the case where dropping it changes the answer.
• "In question 12, BD is the 6 cm." BD is 8 cm and DC is 6 cm; the figure prints 6 cm on the C side. Swapping them changes AB and AC.
• "The tangent length is the distance from the point to the circle." In question 1 the point is 25 cm from the centre of a 7 cm circle — 18 cm from the nearest point of the circle, and 24 cm along the tangent.

Questions to check understanding

• State and prove Theorem 10.2
• Find a radius, a tangent length, or a distance to the centre, given the other two
• Multiple choice on the angle at an external point given the angle at the centre, and the reverse
• Prove that the segment to the centre bisects the angle between the two tangents
• Prove the chord of the larger of two concentric circles touching the smaller is bisected at the touching point, or compute its length
• Find a tangent length from a chord and a radius, by similar triangles or by Pythagoras
• Prove the opposite-sides sum relation for a quadrilateral drawn round a circle, and use it to identify a rhombus
• Find the remaining sides of a triangle drawn round a circle from the inradius and the two parts of one side
• Prove the supplementary relation between the angle at an external point and the angle at the centre

Examples worth working on the board

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

• Fig. 10.7, exactly as drawn (§10.3, p. 149). Read from the printed page: the external point P sits on the left, the centre O on the right, inside the circle. The two touching points are Q above and R below. The two tangent segments PQ and PR are drawn solid; the three joining segments PO, OQ and OR are drawn dashed. The dashing is the drawing's way of saying these are the auxiliary lines the proof adds, and the redraw should keep it and show them in.
• The proof, as inputs (p. 149). Given: a circle centred O, a point P outside it, two tangents from P touching at Q and R. Draw OP, OQ, OR. The angles at Q and at R are right angles, each by Theorem 10.1. Then in the two right triangles: OQ = OR because both are radii; OP is shared. RHS congruence applies, and CPCT gives PQ = PR. Verified as a chain: every one of the four facts used is either given, a radius, shared, or Theorem 10.1 — nothing is read off the figure.
• Why RHS and not SAS. This is added here. Two triangles agreeing on two sides and a non-included angle are not in general congruent — that is the classic ambiguous case. It is the angle being a right angle that removes the ambiguity, which is exactly why the criterion has its own name. So the right angle is not decoration in this proof; it is the whole licence.
• Remark 1, the Pythagoras route (p. 149). Verified: PQ² = OP² − OQ² and PR² = OP² − OR², and since OQ = OR the two right-hand sides are the same number, so PQ = PR. What this route shows and the congruence route hides is the actual value: both lengths are √(OP² − r²), so the equality is not a coincidence of two triangles but the observation that one formula was evaluated twice on the same inputs.
• Remark 2, the free corollary (p. 149). The same congruence also matches the angles at P, so OP splits the angle between the two tangents into two equal halves — the centre lies on that bisector. Almost every later question uses this rather than the equality.
• Example 1, the concentric circles (p. 149–150, Fig. 10.8). Given: circles C₁ and C₂ built on one shared centre O, with C₂ the smaller; the outer circle carries a chord AB which happens to graze C₂, at a point P. To show: P is the midpoint of AB. Read from the printed page: the outer circle is labelled C₁ at its upper right and the inner one C₂ beside the centre; A sits on the left of the outer circle and B on its lower right; P is the touching point on the inner circle; OP is drawn dashed with a right-angle mark at P. Verified as a chain: AB is a tangent to C₂ at P, so Theorem 10.1 makes OP perpendicular to AB; AB is a chord of C₁ and OP runs from its centre; the perpendicular from a centre onto a chord bisects it, so AP = BP. Note the pivot: the same segment AB is a tangent to one circle and a chord of the other, and each role contributes one step.
• Example 2, the two angles (p. 150, Fig. 10.9). Given: a circle centred O, an external point T, tangents TP and TQ touching at P and Q. To show: the angle at T is twice the angle OPQ. Verified on Fig. 10.9: T is on the left, O inside the circle to the right; P is the upper touching point and Q the lower; the segments TP, TQ, the chord PQ and the radius OP are all drawn — OQ is not drawn at all — and there are exactly two marked angles, one at T between the tangents and one at P between PQ and PO. The figure has been stripped to precisely the two quantities in the claim. Verified as algebra: let the angle at T be θ; TP = TQ by Theorem 10.2 so the triangle on T, P, Q is isosceles and each of its base angles is (180° − θ)/2 = 90° − θ/2; the angle OPT is 90° by Theorem 10.1; subtracting gives the angle OPQ = 90° − (90° − θ/2) = θ/2.
• Example 3, a tangent length from a chord (p. 150–151, Fig. 10.10). Given: a chord PQ of length 8 cm in a circle of radius 5 cm, with the tangents at P and at Q meeting at T. Find TP. Read from the printed page: T on the left, O on the right, P at the top of the circle and Q at the bottom, R the point where OT crosses PQ; the labels 5 cm on the radius OP and 8 cm on the chord PQ are printed on the drawing, and TO is dashed. Verified, route one: the triangle on T, P, Q is isosceles and TO bisects its apex angle, so OT is perpendicular to PQ and cuts it in half, giving PR = RQ = 4 cm; then OR = √(5² − 4²) = 3 cm; the right triangles TRP and PRO are similar by AA, so TP/PO = RP/RO, that is TP/5 = 4/3, so TP = 20/3 cm, about 6.67 cm. Verified, route two (the chapter's own note): put TP = x and TR = y; the right triangle on P, R, T gives x² = y² + 16; the right triangle on O, P, T gives x² + 25 = (y + 3)²; subtracting leaves 6y = 32, so y = 16/3, and then x² = 256/9 + 144/9 = 400/9 and x = 20/3. The two routes agree, and the agreement is worth showing — it is the cheapest possible demonstration that the geometry is not route-dependent.
• The four-right-angles shortcut. This is added here as a named move. Join the external point to the centre and to both touching points. The four-sided figure O, P, T, Q has right angles at P and at Q, so the angle at T and the angle at O must total 180°. That single sentence answers Exercise 10.2 questions 2, 3 and 10 outright.
• Exercise 10.2 question 1 (p. 151). Tangent length 24 cm, distance from the point to the centre 25 cm; options 7, 12, 15, 24.5 cm. Verified: radius √(625 − 576) = 7 cm.
• Exercise 10.2 question 2 (p. 151, Fig. 10.11). Tangents TP and TQ with the angle POQ at the centre given as 110°; options 60°, 70°, 80°, 90°. Read from the printed page: T is at the upper right, P at the upper left of the circle, Q at its right, the 110° is marked at O between OP and OQ. Verified by the shortcut: the angle at T is 180° − 110° = 70°.
• Exercise 10.2 question 3 (p. 151). Tangents PA and PB inclined to each other at 80°; find the angle POA; options 50°, 60°, 70°, 80°. Verified: PO bisects, so the angle APO is 40°; the angle at A is 90°; so the angle POA is 50°.
• Exercise 10.2 question 4 (p. 152). Prove the tangents at the two ends of a diameter are parallel. Verified sketch: each is perpendicular to the same line, the diameter itself.
• Exercise 10.2 question 5 (p. 152). Prove the perpendicular erected at the point of contact passes through the centre — the converse of Theorem 10.1. Verified sketch: two distinct perpendiculars to one line at one point cannot both exist.
• Exercise 10.2 question 6 (p. 152). A point 5 cm from the centre, tangent length 4 cm. Verified: radius 3 cm.
• Exercise 10.2 question 7 (p. 152). Concentric circles of radii 5 cm and 3 cm; find the chord of the larger that touches the smaller. Verified: the touching point halves the chord and sits 3 cm from the centre, so each half is √(25 − 9) = 4 cm and the chord is 8 cm. Note: this is Example 1 supplying the reasoning and Example 3's numbers running backwards — the same 5, 4, 3 and 8 appear in all three.
• Exercise 10.2 question 8 (p. 152, Fig. 10.12). A quadrilateral ABCD drawn round a circle; prove AB + CD = AD + BC. Read from the printed page: D upper left, C top right, B lower right, A lower left; the touching points are dotted and labelled R on DC, Q on CB, P on AB and S on DA. Verified: the two tangent lengths from each vertex agree, so AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS), which regroups as (AS + SD) + (BQ + QC) = AD + BC.
• Exercise 10.2 question 9 (p. 152, Fig. 10.13). Two parallel tangents XY and X′Y′, and a third tangent AB touching at C and meeting them at A and at B; prove the angle AOB is 90°. Read from the printed page: XY runs across the top touching at P, X′Y′ across the bottom touching at Q, and AB slants down the right-hand side touching at C, with OA, OB and OC drawn. Verified: OA bisects the angle at A and OB the angle at B; those two angles are co-interior between the parallels so they total 180°; their halves total 90°; and the third angle of the triangle AOB is therefore 90°.
• Exercise 10.2 question 10 (p. 152). Prove the angle between the two tangents from an external point is supplementary to the angle the segment joining the touching points subtends at the centre. Verified: this is the four-right-angles shortcut stated in full generality, and question 2 is its numerical instance.
• Exercise 10.2 question 11 (p. 152). Prove a parallelogram drawn round a circle is a rhombus. Verified: question 8 gives AB + CD = AD + BC; in a parallelogram AB = CD and AD = BC, so 2·AB = 2·AD and all four sides agree.
• Exercise 10.2 question 12 (p. 152, Fig. 10.14). A triangle ABC drawn round a circle of radius 4 cm, with the touching point D dividing BC into BD = 8 cm and DC = 6 cm; find AB and AC. Read from the printed page: A at the apex, C at the lower left, B at the lower right, the incircle centred O, and D on BC with a 6 cm arrow from C to D and an 8 cm arrow from D to B — so the printed figure puts the 6 on the C side, matching the question's ordering of BD then DC. Verified: write the equal tangent lengths from A as x; then BC = 14, CA = x + 6, AB = x + 8, and the half-perimeter is x + 14. Area by the incircle is 4(x + 14); area by Heron is √((x + 14)·x·8·6). Squaring and cancelling the common factor gives 16(x + 14) = 48x, so x = 7, and AB = 15 cm, AC = 13 cm. Check: sides 13, 14, 15 have half-perimeter 21 and area √(21·7·8·6) = 84, so the inradius is 84/21 = 4 cm as given.
• Exercise 10.2 question 13 (p. 152). Take a quadrilateral drawn round a circle, pick two sides facing each other, and show that the angles they subtend at the centre add to 180°. Verified sketch: the eight small angles at the centre pair off equal by the same congruences that give the equal tangent lengths, and the eight total 360°, so the two opposite-side angles total 180°.

Figures to have open

• Fig. 10.7 redrawn (p. 149) as the book draws it — P on the left, O at the centre on the right, Q above and R below in mirrored positions, PQ and PR solid and PO, OQ and OR all dashed. The chapter's own figure.
• The two right triangles OQP and ORP pulled apart and set side by side, with matching parts colour-coded. Standard schematic, not in the book — the chapter compares them in words only.
• Fig. 10.8, the concentric pair (p. 149), with AB drawn once and labelled twice, as a tangent and as a chord. The chapter's own figure.
• Fig. 10.9 (p. 150), which is already stripped to the two angles in the claim and should be redrawn as it stands, including the fact that only one radius is drawn.
• Fig. 10.10 (p. 150), carrying the printed 5 cm and 8 cm and the point R where OT meets the chord. The chapter's own figure.
• Fig. 10.12 (p. 152), the quadrilateral round a circle with the four touching points P, Q, R, S dotted, for the adding-up argument. The chapter's own figure.
• Fig. 10.14 (p. 152), the triangle round a circle with 6 cm and 8 cm marked on the two parts of BC. The chapter's own figure; keep 6 cm on the C side as printed.
• A colour-coding scheme for equal tangent pairs, reused across the quadrilateral and the triangle. Standard schematic, not in the book — it is the only thing that makes questions 8, 11, 12 and 13 look like one idea rather than four.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 10 "Circles", §10.3, pp. 149–151: the statement and proof of Theorem 10.2 with Fig. 10.7, Remarks 1 and 2, and Examples 1, 2 and 3 with Figs. 10.8, 10.9 and 10.10
• Exercise 10.2, questions 1 to 13, pp. 151–152, with Figs. 10.11, 10.12, 10.13 and 10.14
• Theorem 10.1 and Remark 1, p. 147, which supply every right angle used above
• The definition of the length of a tangent, §10.3, p. 148
• The chapter summary, §10.4, p. 153, whose third point is Theorem 10.2
• The Class IX result that the perpendicular from the centre bisects a chord is used in Example 1 and is not restated in this chapter