Tangents Class 10 ICSE — Bansal Chapter 18
Tangents class 10 ICSE covers the geometry of lines that touch a circle at exactly one point. This chapter in the Bansal Mathematics book explains the tangent radius theorem, common tangents, intersecting chords, and the alternate segment theorem. You will learn when two circles touch and how to find tangent lengths. Worked examples show the method for each result.
Download Tangents and Intersecting Chords Class 10 ICSE Mathematics Chapter 18 PDF
Download the Tangents and Intersecting Chords Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 20 pages, 1 sections, 66 questions.
What is inside Chapter 18
20 pages · 1 sections · 12 worked examples · 66 exercise questions
| Section | Title |
|---|---|
| 18.1 | Introduction |
- EXERCISE 18(A): 23 questions
- EXERCISE 18(C): 43 questions
18.1 Introduction
Definitions: three positions of a line and a circle
Draw a line and a circle on the same plane. The line can sit in one of three positions relative to the circle. First, the line may not meet the circle at all. Second, it may cut the circle at two points. Third, it may touch the circle at exactly one point.
A line that cuts a circle at two points is called a secant of the circle. A line that touches the circle at one point only is called a tangent. The point where a tangent touches the circle is called the point of contact.
Three facts decide how many tangents you can draw from a given point. No tangent can pass through a point inside the circle. Exactly one tangent passes through a point on the circumference. Exactly two tangents pass through a point outside the circle. All the theorems in this tangents class 10 ICSE chapter build on these three positions.
Theorem 10: tangent perpendicular to the radius
Theorem 10 states that the tangent at any point of a circle is perpendicular to the radius through that point. If AB is a tangent at point P and OP is the radius, then OP \perp AB. The proof uses a shortest-segment idea. Take any other point Q on the tangent. Because Q lies outside the circle, OQ is longer than OP.
So OP is the shortest segment from O to the tangent line. The shortest segment from a point to a line is always perpendicular to that line. Hence OP is perpendicular to AB.
The corollary gives three results when two tangents PA and PB are drawn from an exterior point P. First, PA = PB. Second, the tangents subtend equal angles at the centre, so \angle AOP = \angle BOP. Third, the tangents are equally inclined to the line joining P to the centre, so \angle APO = \angle BPO. The proof compares triangles AOP and BOP.
OA = OB as radii, \angle OAP = \angle OBP = 90^\circ, and OP is common, so the triangles are congruent by RHS.
Theorem 11: two touching circles
Theorem 11 says that when two circles touch, the point of contact lies on the straight line through the centres. The proof draws a common tangent at the point of contact. Each radius is perpendicular to that tangent. Two perpendiculars at the same point must lie on the same line, so the centres and the point of contact are collinear.
The distance between centres follows a simple rule. If two circles touch externally, the distance is the sum of the radii, d = r_1 + r_2. If they touch internally, the distance is the difference, d = r_1 - r_2, taking the larger radius first.
Two more facts are worth remembering. If two parallel tangents touch a circle at points P and Q, then PQ is always a diameter of that circle. Concentric circles are circles with the same centre. This theorem appears often with two touching circles and a common tangent drawn through the point of contact.
Direct and transverse common tangents
A common tangent touches two circles. When both circles lie on the same side of the tangent, it is called a direct common tangent. When the circles lie on opposite sides of the tangent, it is called a transverse common tangent.
To find the length of a direct common tangent, use a rectangle construction. Suppose the centres are A and B, with radii 12 cm and 5 cm, and the centres are 17 cm apart. The distance AB = 17 cm. Drop a perpendicular BC to the radius AM. Then AC = 12 - 5 = 7 cm. In right triangle ABC, AB^2 = AC^2 + BC^2, so BC^2 = 289 - 49 = 240.
Hence BC = \sqrt{240} = 4\sqrt{15} cm. The tangent length equals BC because BCMN is a rectangle.
| Feature | Direct common tangent | Transverse common tangent |
|---|---|---|
| Position of circles | Both on the same side | On opposite sides |
| Horizontal gap | r_1 - r_2 | r_1 + r_2 |
| Length | \sqrt{d^2 - (r_1-r_2)^2} | \sqrt{d^2 - (r_1+r_2)^2} |
Look at which side each circle lies on to decide which formula applies. The direct tangent uses the difference of radii. The transverse tangent uses the sum. Pairing the wrong formula is a common error in this part of the chapter.
Theorem 12: intersecting chords
Theorem 12 deals with chords that intersect. If two chords AB and CD of a circle intersect at a point P, then the products of the segments are equal: PA \times PB = PC \times PD. The result holds whether the chords cross inside the circle or meet outside it when produced.
The proof joins the endpoints to form triangles. In the internal case, triangles APC and BPD are similar by the AA postulate, because angles A and D stand on the same segment and angles C and B also stand on the same segment. Corresponding sides give the proportion \frac{PA}{PD} = \frac{PC}{PB}, which rearranges to the product.
In the external case, triangles PAC and PDB are similar using the exterior angle of a cyclic quadrilateral. The same proportion follows, and the same product equality. These proofs lean on the similarity of triangles, covered earlier in the Class 10 Maths book.
The most common error is pairing the wrong segments. The product always pairs the two segments of one chord against the two segments of the other. Mark each chord’s two pieces before substituting numbers.
Theorem 13: alternate segment
Theorem 13, called the alternate segment theorem, links a tangent and a chord. The angle between a tangent and a chord through the point of contact equals the angle in the alternate segment.
Suppose tangent PQ touches the circle at point A, and chord AB is drawn through A. If C lies on the major arc AB, then \angle BAQ = \angle ACB. If D lies on the minor arc AB, then \angle BAP = \angle ADB.
The proof draws diameter AOR and joins RB. In triangle ABR, \angle ABR = 90^\circ because it stands on the diameter. Since OAQ is a right angle, the two right-angle relations together show that \angle ARB = \angle BAQ. Then \angle ARB = \angle ACB because both stand on the same segment. The second equality follows from the straight angle at A and the opposite angles of the cyclic quadrilateral.
Most of the later exercises in this ICSE chapter use this theorem once, then combine it with the angle-sum of a triangle to reach the result.
Theorem 14: tangent and chord
Theorem 14 combines a chord and a tangent that meet outside the circle. If chord AB and tangent TP intersect at point P outside the circle, then PA \times PB = PT^2. The square of the tangent length equals the product of the chord segments.
The proof joins TA and TB to form triangles PAT and PTB. These triangles are similar by the AA postulate. Angle PTB equals angle A by the alternate segment theorem, and angle P is common. The proportion \frac{PA}{PT} = \frac{PT}{PB} rearranges to PA \times PB = PT^2.
Here is a worked pattern with fresh numbers. Let PA = 9 cm and PB = 4 cm. Then PT^2 = 9 \times 4 = 36, so PT = 6 cm. The tangent length is the geometric mean of the two chord segments.
A common error is using the whole chord as one segment. The segment PB runs from the external point to the nearer end of the chord, not the full chord length.
Worked examples with original numbers
Example 1: Inscribed circle in a right triangle. Let triangle PQR have angle Q = 90°, with PQ = 8 cm and QR = 6 cm. The hypotenuse is PR = \sqrt{64 + 36} = 10 cm. Let the inscribed circle have radius x. The square at the right angle has side x, so PA = 8 - x and RB = 6 - x.
Equal tangent segments give PC = 8 - x and RC = 6 - x. Since PR = PC + RC, we get 10 = (8-x) + (6-x), so 2x = 4 and x = 2. The radius is 2 cm.
Example 2: Direct common tangent. Two circles with radii 12 cm and 5 cm touch externally, with centres 17 cm apart. The tangent length is \sqrt{17^2 - 7^2} = \sqrt{240} = 4\sqrt{15} cm, as shown earlier.
Example 3: Intersecting chords. Chords AB and CD meet inside the circle at P. If PA = 3 cm, PB = 8 cm, and PC = 4 cm, then 3 \times 8 = 4 \times PD, giving PD = 6 cm. Set up the product, substitute, and solve.
What to practise from this chapter
Two exercise blocks carry the practice for this chapter in the Class 10 ICSE Maths book:
- EXERCISE 18(A) — 23 questions on tangents, the tangent radius theorem, equal tangents from an exterior point, and two touching circles.
- EXERCISE 18(C) — 43 questions covering intersecting chords, the alternate segment theorem, and tangent-chord problems.
Many questions in both exercises carry ICSE exam years from 2000 to 2015. For example, the inscribed-circle question carries 2012, and the direct common tangent question carries 2004. The past years marked on the questions show this tangents class 10 ICSE chapter is a regular board topic.
The chapter draws on earlier work on circles and similarity of triangles. Revise the angle properties of a circle before attempting the exercises. If you need the full set of Class 10 Maths books, visit the Class 10 Maths hub, or browse all Class 10 books. For the official ICSE Mathematics syllabus and examination details, see the CISCE website.
Frequently asked questions
What is the tangent radius theorem in tangents class 10 ICSE?
The tangent radius theorem says the tangent at any point of a circle is perpendicular to the radius through that point. It holds because every other point on the tangent lies outside the circle, so the radius is the shortest segment from the centre to the tangent. Use it to set up right angles whenever a tangent touches a circle.
How do I use the intersecting chords theorem to find an unknown segment?
Write the product PA \times PB = PC \times PD, pairing the two segments of one chord against the two segments of the other. Substitute the three known lengths and solve for the unknown. The same product works whether the chords meet inside or outside the circle.
What is the alternate segment theorem and when do I apply it?
The alternate segment theorem says the angle between a tangent and a chord through the point of contact equals the angle in the alternate segment. Apply it whenever a tangent and a chord meet at a point on the circle. It is the main tool in the later exercises of this ICSE chapter.
What is the difference between a direct and a transverse common tangent?
A direct common tangent keeps both circles on the same side of the tangent. A transverse common tangent has the circles on opposite sides. The length formula uses the difference of radii for a direct tangent and the sum of radii for a transverse tangent.
How many tangents can be drawn from a point outside a circle?
Exactly two tangents can be drawn from a point outside a circle. No tangent can pass through a point inside the circle, and exactly one tangent passes through a point on the circumference.
Reference: ICSE Class 10 Mathematics textbook material, chapter 18: Tangents and Intersecting Chords — CISCE-aligned.
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