Circles Class 10 ICSE — A Das Gupta Chapter 21

Written by Greya LakshmiPublished on 28 September 2026

Circles Class 10 ICSE — Chapter 21 of A. Das Gupta’s Mathematics — covers the tangent properties of a circle. The chapter defines the segments of a chord and the alternate segments of a circle, then proves three theorems: intersecting chords, the tangent–secant rule, and the alternate segment theorem. Each theorem ends in a product or angle equality you can apply directly to a figure.

The sections below follow the book’s order, with the formulas and a worked pattern for each.

Download Tangent Properties of Circles Class 10 ICSE Mathematics Chapter 21 PDF

Download the Tangent Properties of Circles Class 10 ICSE Mathematics chapter PDF from the A. Das Gupta textbook to revise offline or print for practice. It is the full chapter as printed in the book — 22 pages, 3 sections, 44 questions.

What is inside Chapter 21

22 pages · 3 sections · 44 exercise questions

Section	Title
1	Intersecting chords and Tangents
2	Alternate Segments
3	Some more Results (Theorems)

EXERCISE 21 (A): 27 questions
EXERCISE 21 (B): 17 questions

1 Intersecting chords and Tangents

When a point P lies on a chord AB of a circle, P divides AB internally into two segments, PA and PB. When AB is produced past an endpoint and P lies on that extended line, P divides AB externally into two segments. The segment names stay PA and PB in both cases; only the position of P changes.

Suppose chord AB is $10\ \text{cm}$ long and P is $4\ \text{cm}$ from A. Internally, the segments are $PA = 4\ \text{cm}$ and $PB = 6\ \text{cm}$ , which add to the chord. Externally, P lies beyond the chord, so one segment contains the other and the sum no longer holds.

This shared notation lets Theorem 1 state one product rule, $PA \times PB = PC \times PD$ , for both positions of P.

The same segment language reappears in the angle properties of a circle chapter, where angles in the same segment do similar work. In this Class 10 ICSE chapter, deciding which length is the near segment and which is the far segment is the first step of every chord problem.

2 Alternate Segments

In the book’s figure, PAT is a tangent to the circle at A and AB is a chord drawn from A. The chord divides the circle into two regions. These regions are the alternate segments: one is ADB and the other is BCA.

The mapping fixes which angle lives in which region:

For $\angle BAT$ , the alternate segment is BCA.
For $\angle BAP$ , the alternate segment is ADB.

Why does the name matter? Theorem 3 states that the angle between the tangent and the chord equals an angle in the alternate segment, not in the adjacent one. Choose the wrong region and the equality fails. Memory aid: follow the angle from the tangent line across the chord; the region it enters is the alternate segment.

Angle BAT is formed by the tangent ray AT and the chord AB, so it opens into the region BCA. Angle BAP uses the other ray of the tangent and opens into ADB. Naming the segment correctly is the step students skip, and the error shows up in the final answer.

3 Some more Results (Theorems)

Theorem 1. If two chords of a circle intersect internally or externally, the products of the lengths of their segments are equal.

Case 1 — internal. Chords AB and CD cross at P inside the circle. Join AC and BD. In triangles APC and DPB, angle APC equals angle DPB because they are vertically opposite, and angle PAC equals angle PDB because both stand on the same arc. So $\Delta APC \sim \Delta DPB$ by the AA axiom, and corresponding sides give:

$\frac{PA}{PD} = \frac{PC}{PB} \Rightarrow PA \times PB = PC \times PD$

Case 2 — external. The chords, produced, meet at P outside the circle. Here triangles PDB and PAC are similar. The equal angles are exterior angles of the cyclic quadrilateral ACBD: $\angle PDB = \angle PAC$ and $\angle PBD = \angle PCA$ . Corresponding sides give:

$\frac{PD}{PA} = \frac{PB}{PC} \Rightarrow PA \times PB = PC \times PD$

Why does one formula cover both drawings? In each case P is the common vertex of two secant lines, and AA similarity converts the ratio of matching sides into the same cross-product. The figure changes; the algebra does not.

Internal versus external intersection.

Feature	Internal intersection	External intersection
Position of P	Inside the circle	Outside the circle
How the chords meet	AB and CD cross at P	AB and CD produced meet at P
Equal angles used	Vertically opposite; same segment	Exterior angle of cyclic quadrilateral
Similar triangles	$\Delta APC \sim \Delta DPB$	$\Delta PDB \sim \Delta PAC$
Product result	$PA \times PB = PC \times PD$	$PA \times PB = PC \times PD$

Theorem 2. If a chord and a tangent intersect externally, the product of the chord’s segments equals the square of the tangent length.

The figure has a tangent touching the circle at T and a chord AB produced to meet it at P. The claim is $PA \times PB = PT^2$ . The proof draws $OL \perp AB$ , then joins OB, OP, and OT.

Three facts drive the proof. The perpendicular from the centre bisects the chord, so $AL = LB$ . The radius through the point of contact is perpendicular to the tangent, so $OT \perp TP$ . Pythagoras then links the segments:

$PA \times PB = (PL + AL)(PL - LB) = (PL + LB)(PL - LB) = PL^2 - LB^2$

From right triangle OLP, $PL^2 = OP^2 - OL^2$ . From right triangle OLB, $OL^2 + LB^2 = OB^2$ . So the product becomes $OP^2 - OB^2$ . Since OB and OT are radii, $OB = OT$ , and triangle OTP is right-angled, so $OP^2 - OT^2 = PT^2$ . Hence $PA \times PB = PT^2$ .

Worked example with new numbers. A tangent PT of length $15\ \text{cm}$ is drawn from an external point P to a circle. A secant through P cuts the circle at A and B, with A nearer to P, and $PB = 25\ \text{cm}$ .

Step 1: Apply the tangent–secant theorem $PT^2 = PA \times PB$ .

$15^2 = PA \times 25 \Rightarrow 225 = 25 \times PA \Rightarrow PA = 9\ \text{cm}$

Step 2: Subtract the near segment from the full secant to get the chord: $AB = PB - PA$ .

$AB = 25 - 9 = 16\ \text{cm}$

Final answer: $PA = 9\ \text{cm}$ and $AB = 16\ \text{cm}$ .

Theorem 3. The angle between a tangent and a chord through the point of contact equals an angle in the alternate segment.

With PAT the tangent at A and AB the chord, the theorem gives two equalities:

$\angle BAT = \angle ACB \quad \text{and} \quad \angle BAP = \angle ADB$

Proof idea: draw the diameter AOE and join EB. Angle ABE is $90^\circ$ because it stands in a semicircle, so $\angle AEB + \angle EAB = 90^\circ$ . The radius through the point of contact is perpendicular to the tangent, so $\angle EAT = 90^\circ$ , meaning $\angle EAB + \angle BAT = 90^\circ$ .

Comparing the two sums gives $\angle AEB = \angle BAT$ . But $\angle AEB$ and $\angle ACB$ stand on the same arc AB, so they are equal. Hence $\angle BAT = \angle ACB$ .

The second equality follows from the first. Angles BAP and BAT form a straight line, so $\angle BAP + \angle BAT = 180^\circ$ . Also, $\angle ADB + \angle ACB = 180^\circ$ because ADBC is a cyclic quadrilateral. Replacing $\angle BAT$ with $\angle ACB$ gives $\angle BAP = \angle ADB$ .

Memory aid for the alternate segment theorem: the angle between the tangent and the chord equals the angle in the segment on the far side of the chord — the region that does not touch the tangent.

Tangent length by Pythagoras. The radius–tangent right angle also gives tangent lengths directly. For a tangent AP touching a circle of radius OA, $OA \perp AP$ , so $OP^2 = OA^2 + AP^2$ . One known side pair is enough to find the third.

Common mistakes and corrections.

Using the wrong segment product. Write $PA \times PB$ , never $PA \times AB$ . Both factors are segments measured from P — the near segment and the far segment.
Forgetting the radius–tangent right angle. The radius through the point of contact is perpendicular to the tangent. That $90^\circ$ angle starts every Pythagoras application.
Mixing the internal and external cases. Internally, P is inside the circle and the chords cross. Externally, P is outside and the chords are produced to meet. The product formula is the same, but the similar triangles in the proof are different.

Board exam pattern. Past board questions appear inside Exercise 21 (B) with year labels 2006, 2007, 2008, and 2009. In the Class 10 ICSE board exam, three patterns recur: chord products, tangent–secant products, and alternate segment angle equalities. Many questions combine two of them in one figure, such as a tangent with a diameter, or chords meeting at right angles.

What to practise from Circles Class 10 ICSE

The chapter has two exercise blocks. Practise both in order; the first builds the tangent facts, the second applies all three theorems in questions that include past board problems.

EXERCISE 21 (A) — 27 questions
EXERCISE 21 (B) — 17 questions, including past board questions labelled 2006, 2007, 2008, and 2009

For the latest syllabus and exam format, check the official CISCE website. To see where circle geometry fits, browse the Class 10 ICSE Maths book list or the wider Class 10 ICSE book collection.

Circles Class 10 ICSE frequently asked questions

What is the tangent secant theorem in circles class 10 ICSE?

It states that when a chord and a tangent intersect externally at a point P, the product of the chord’s two segments equals the square of the tangent length: $PA \times PB = PT^2$ . The proof drops a perpendicular from the centre to the chord, then applies Pythagoras to three right triangles.

How do I use the chord product rule when two chords meet inside a circle in Class 10 ICSE?

Take the two chords $AB$ and $CD$ meeting at P. Multiply the two segments of the first chord, $PA$ and $PB$ , and set the product equal to the product of the two segments of the second chord, $PC$ and $PD$ . Substitute the three known lengths and solve for the fourth. The same formula applies when the chords are produced to meet outside the circle.

What does the alternate segment theorem state in this ICSE chapter?

It states that the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment. In the book’s notation, $\angle BAT = \angle ACB$ and $\angle BAP = \angle ADB$ . The alternate segment is the region on the far side of the chord, away from the tangent.

How do I find the length of a tangent drawn from an external point in Class 10 ICSE?

Join the centre to the point of contact. The radius is perpendicular to the tangent, so the radius, the tangent, and the line from the centre to the external point form a right triangle. Use Pythagoras: $AP^2 = OP^2 - OA^2$ . For example, with $OP = 13\ \text{cm}$ and $OA = 5\ \text{cm}$ , $AP = \sqrt{169 - 25} = 12\ \text{cm}$ .

Which exercises in this chapter should I practise for the ICSE Class 10 board exam?

Practise both blocks. Exercise 21 (A) builds the basic tangent and chord skills with 27 questions, and Exercise 21 (B) applies all three theorems in 17 questions, including past board questions labelled 2006 to 2009. The chord product, tangent–secant product, and alternate segment angle patterns are the ones that recur. Other Class 10 ICSE study pages cover the chapters that build on these ideas.

Reference: ICSE Class 10 Mathematics textbook material, chapter 21: Tangent Properties of Circles — CISCE-aligned.

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