Tangents and Secants to a Circle
1. What This Chapter Covers
Two lines in a plane can intersect, stay apart, or coincide. Ask the same question of a curve and a line, and the answer depends on the curve — but for a circle the possibilities are few enough to list.
Abhiram argues there can be only three. Draw a circle with centre O and a line PQ, and either they have no point in common, or two, or exactly one.
A line meeting the circle at two points is a secant, and the piece of it between those points is a chord. A line meeting the circle at exactly one point is a tangent, and that point is the point of contact.
The book adds a nice piece of etymology: tangent comes from the Latin tangere, to touch, and the term was introduced in 1583 by a Danish mathematician the page calls "Thomas Fineke" — almost certainly a misprint for Thomas Fincke.
The index allots this chapter 15 periods in November, and it runs from textbook page 229 to page 248, with three numbered exercises and an Optional Exercise.
2. A Tangent Is a Secant Whose Two Points Have Merged
An activity makes the relationship clear. Attach a straight wire AB to a circular wire at a point P so that it can pivot, and rotate it. In almost every position the straight wire crosses the circle at P and at a second point — Q₁, Q₂, Q₃ as it turns. In one position only, the second point catches up with P.
So the book defines the tangent as a limiting case: a tangent is a special case of a secant where the two points of intersection of a line with a circle coincide.
That is worth holding on to, because it explains why there is exactly one tangent at a point rather than two or none. The pivoting wire passes through the tangent position once, and once only.
3. The Tangent Is Perpendicular to the Radius
Theorem 9.1 is the chapter's foundation: the tangent at any point of a circle is perpendicular to the radius through the point of contact.
The proof is short and entirely about distances. Take any point Q on the tangent other than P. Q must lie outside the circle — if it were inside, the line would cut the circle twice and be a secant, not a tangent. So OQ is longer than the radius, which is OP.
That holds for every point of the line, so OP is the shortest of all the distances from O to the line. And the shortest segment from a point to a line is the perpendicular. Therefore OP ⊥ XY.
Two consequences follow immediately. Since there is only one perpendicular to OP at P, one and only one tangent can be drawn at a given point of a circle. And since there is only one perpendicular to the tangent at P, the perpendicular to a tangent at its point of contact passes through the centre.
The book adds a naming note: the line containing the radius through the point of contact is also called the normal to the circle at that point.
4. How Many Tangents from a Point
Where the point sits decides the answer, and there are only three cases.
If P is inside, every line through it cuts the circle twice, so all of them are secants and none is a tangent. If P is on the circle, there is exactly one. If P is outside, there are exactly two, and their points of contact are the two places the circle turns away from P.
For an external point, the segment from P to a point of contact is called the length of the tangent from P. Because the tangent meets the radius at a right angle, that length is found by Pythagoras: with radius 6 cm and OP = 10 cm, the tangent length is √(100 − 36) = 8 cm.
The reason the inside case gives nothing is worth stating plainly. A tangent touches the circle without entering it, and a line through an interior point has already entered. So every such line must leave again, meeting the circle a second time, and every line through an interior point is a secant.
The book reaches the same three cases by experiment rather than argument, in an Activity that asks students to take a point inside, then on, then outside, and try in each case. The count 0, 1, 2 is then observed before it is explained.
5. The Two Tangent Lengths Are Equal
Theorem 9.2 says the lengths of tangents drawn from an external point to a circle are equal.
The proof needs only the previous theorem. Both angles OAP and OBP are 90°, the radii OA and OB are equal, and OP is common to both triangles — so triangles OAP and OBP are congruent by RHS, and PA = PB follows from corresponding parts.
The chapter then draws out four consequences, each stated as a separate result with a proof.
The centre lies on the bisector of the angle between the two tangents. The same congruence gives ∠QPO = ∠OPR, so OP bisects ∠QPR.
In two concentric circles, a chord of the larger circle that touches the smaller one is bisected at the point of contact. Since that chord is a tangent to the inner circle, the radius to the point of contact is perpendicular to it — and a perpendicular from the centre bisects a chord.
If two tangents AP and AQ are drawn from an external point A, then ∠QAP = 2∠QPO = 2∠OQP. Writing ∠QAP as θ, the isosceles triangle APQ gives ∠APQ = 90° − θ/2, and subtracting that from the right angle ∠OPA leaves θ/2.
A quadrilateral drawn around a circle satisfies AB + CD = BC + DA.
The proof is just Theorem 9.2 applied at each of the four vertices, then added. Grouping the eight equal pieces into the four sides gives the result, and Exercise 9.2 uses it to prove that a parallelogram circumscribing a circle must be a rhombus.
6. Constructions
The chapter gives two constructions and justifies both.
A tangent at a given point on the circle. Join the centre to the point and draw the perpendicular there. Theorem 9.1 guarantees that this perpendicular is the tangent, and that there is no second one.
The two tangents from an external point P. Join PO, bisect it at M, and draw a circle with centre M and radius MP. It cuts the given circle at A and B, and PA and PB are the tangents.
The justification is the semicircle theorem. Since PO is a diameter of the circle centred at M, the angle PAO standing on it is 90°. So PA is perpendicular to the radius OA at a point of the circle, and by the converse of Theorem 9.1 it must be a tangent.
Example-1 is a construction that needs a calculation first. To draw two tangents inclined at 60° to a circle of radius 5 cm, you need the distance of the external point, which the book finds with trigonometry: OP bisects the 60° angle, so ∠OPA = 30°, and sin 30° = OA/OP gives OP = 10 cm.
A Try This box offers a second route to the same construction that avoids trigonometry altogether. Draw two radii OA and OB with ∠BOA = 120°, then draw perpendiculars to OA at A and to OB at B. They meet on the bisector of ∠BOA, and that meeting point is the external point, with the two perpendiculars as the required tangents.
The two methods agree because the angles of the quadrilateral OAPB must total 360°. With right angles at A and B, the angle at O and the angle at P are supplementary — so tangents inclined at 60° correspond to radii at 120°, exactly as the box says. That relationship is the Optional Exercise's first question, stated in general.
7. Segments, and Their Area
Section 9.4 changes subject. A secant cuts a circle into two pieces, each bounded by an arc and a chord, and each is called a segment.
Finding the area needs the sector first. A full circle of radius r has area πr² and 360° at the centre, so a sector of angle x° has area (x/360) × πr².
The minor segment is then just the sector with its triangle taken away:
area of segment APB = (x/360) × πr² − area of ΔOAB
Example-1 works it through for r = 21 cm and x = 120°. The sector is 462 cm². For the triangle, dropping OM perpendicular to AB halves the angle, so cos 60° = OM/21 gives OM = 10.5 and sin 60° = AM/21 gives AM = 21√3/2, making AB = 21√3 and the triangle (441/4)√3. The segment is 462 − 190.95 = 271.05 cm².
The major segment needs no new formula — subtract the minor one from the whole circle.
Example-2 combines the idea with the semicircle theorem. With PQ = 24, PR = 7 and QR a diameter, the angle at P is 90°, so QR = 25 and the radius is 12.5. The two shaded segments together are the semicircle minus the triangle: 245.53 − 84 = 161.53 cm².
Example-3 is a design problem. A round table top of radius 14 cm carries six equal designs, and since the side of a regular hexagon equals the radius of its circumscribing circle, the designs are the circle minus the hexagon: 616 − 509.21 = 106.79 cm², costing Rs 533.95 at Rs 5 per cm².
Before any of that, the chapter pauses on something easy to skip: how to see a composite shape as a sum of familiar ones. Shankar's washbasin picture is a rectangle with a segment on top; a pencil is a rectangle with a triangle; an envelope is a rectangle with two arcs cut out of it.
The instruction the book gives is the whole method: to find the area of a figure, identify what shapes are involved in it. Every question in Exercise 9.3 is that instruction applied — a wiper is a sector, a four-petal design is four differences between a quarter circle and a triangle, and the region between two concentric arcs is one sector minus another.
A Do This box drills the sector formula first, asking for the area of a sector of radius 7 cm at 60°, 30°, 72°, 90° and 120°, and then for the area swept by a 14 cm minute hand in 10 minutes — which is a 60° sector, because ten minutes is a sixth of the dial.
8. What the Exercises Ask, and What the Book Answers
Exercise 9.1 has five questions on the basic vocabulary and two short Pythagoras calculations. Exercise 9.2 has nine, five of them multiple choice and four constructions. Exercise 9.3 has eight on areas of segments and composite shaded regions, and the Optional Exercise adds seven.
The printed answers sit at textbook page 383. Every one of them is correct, which makes this the second chapter in a row with a clean key.
Worth checking your own work against: a tangent of 12 cm from a point 13 cm away on a circle of radius 5, and the same 12 cm from 15 cm away on a radius of 9.
Also a chord of 8 cm in the larger of two concentric circles of radii 5 and 3; a tangent of 2√5 cm from a point on a circle of radius 6 to a concentric circle of radius 4; and the triangle circumscribing a circle of radius 3 with BD = 9 and DC = 3, which gives AB = 15 cm and AC = 9 cm.
The area answers repay a careful look, because they show how varied the shapes get. A chord subtending a right angle in a circle of radius 10 gives a minor segment of 28.5 cm² and a major one of 285.5 cm². A 120° chord in a circle of radius 12 gives 88.368 cm², and two car wipers of blade 25 cm sweeping 115° clean 1254.96 cm² between them.
Two of them land on the same number by different routes. A square of side 10 with semicircles drawn on every side gives a four-petal region of 57 cm²; two quadrants of radius 10 overlapping inside a square give 57 cm² as well. Both reduce to twice the difference between a quarter circle and a right triangle.
9. Three Things in the Chapter Text
The first is in the statement of the chapter's own first theorem. Page 231 sets out Theorem 9.1 like this:
Given: A circle with centre O and a tangent XY to the circle at a point P and OP radius. To prove: OP is perpendicular to XY. i.e XY is tangent to circle.
The clause after "i.e." is not another way of saying OP ⊥ XY. It restates what was given — that XY is a tangent — as though it were the goal. What is to be proved is OP ⊥ XY, and nothing more. The proof that follows does exactly that and never mentions the stray clause, so the slip is in the framing rather than the argument.
The second is the name on page 230. The 1583 introduction of the word tangent is credited to a Danish mathematician printed as "Thomas Fineke", where the usual spelling is Fincke.
The third is a rounding that does not carry through. In Example-3 on page 246 the area of the regular hexagon is printed as 509.2 cm² on the line labelled (2), and then subtracted as 509.21 on the very next line. The exact value is 294√3, which is 509.208, so 509.21 is the better figure and the final answers are right — but the two lines disagree with each other.
Smaller slips run through the chapter: "prependicular" on page 232, "ln ΔOAP" for "In" on page 239, "isoscecles" on page 237, "conœntric" on page 247 and "segmants" on page 248.
10. Summary
A line and a circle can share no point, two points or exactly one. Two points make it a secant, with the piece between them a chord; one point makes it a tangent, and that point is the point of contact. A tangent is the limiting case of a secant whose two intersections have merged.
The tangent at any point is perpendicular to the radius there, and the proof is about distances: every other point of the tangent lies outside the circle, so the radius is the shortest segment from the centre to the line.
A point inside a circle admits no tangent, a point on it exactly one, and a point outside exactly two. For an external point the two tangent lengths are equal, proved by RHS congruence, and the centre lies on the bisector of the angle between them.
A quadrilateral drawn around a circle has AB + CD = BC + DA, from which a circumscribing parallelogram must be a rhombus. A chord of a larger circle touching a smaller concentric one is bisected at the point of contact.
A sector of angle x° has area (x/360) × πr², and the segment cut off by the chord is that sector minus the triangle joining the centre to the chord's ends. The major segment is the whole circle minus the minor one.
