Circles Class 10 ICSE — Bansal Chapter 19
Circles Class 10 ICSE covers the ruler-and-compass constructions of tangents, circumscribed circles, and inscribed circles. Chapter 19 of the Bansal Mathematics book teaches each construction step by step, with the property that makes it work. You learn two ways to draw tangents: through a point on the circumference, and from an exterior point. You then construct circumscribing and inscribing circles for triangles and regular hexagons. You can verify the construction syllabus on the official CISCE website.
Download Constructions Circles Class 10 ICSE Mathematics Chapter 19 PDF
Download the Constructions Circles 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 — 5 pages, 3 sections, 22 questions.
What is inside Chapter 19
5 pages · 3 sections · 22 exercise questions
| Section | Title |
|---|---|
| 19.1 | Construction of Tangents to a Given Circle |
| 19.2 | Construction of Circumscribed and Inscribed Circles of a Triangle |
| 19.3 | Circumscribing and Inscribing a Circle on Regular Hexagon |
- EXERCISE 19: 22 questions
19.1 Construction of Tangents to a Given Circle
Construction 1 draws a tangent through a point on the circumference. Join the centre O to the point P, then draw the line APB through P so that \angle OPA = 90^\circ. The line APB is the required tangent. The rule behind it: the angle between the radius and the tangent at the point of contact is 90^\circ.
Construction 2 draws tangents from an exterior point P. Join P to O, then draw a circle with OP as its diameter. That circle cuts the given circle at A and B. Join PA and PB. Both are tangents to the given circle.
Two properties follow. Tangents drawn from an exterior point are equal in length, so PA = PB. Since the angle of a semi-circle is 90^\circ, \angle PAO = 90^\circ. Pythagoras then gives PA^2 + OA^2 = OP^2, so PA = \sqrt{OP^2 - OA^2} = PB.
Worked example. A circle has radius 3 cm. An exterior point P is 5 cm from the centre O. Then PA = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 cm. Each tangent from P is 4 cm long.
The common mistake is drawing the tangent before joining the radius to the point of contact. Set the 90^\circ angle against the radius already drawn. The same tangent properties appear in the chapter on tangents and intersecting chords in this book.
19.2 Construction of Circumscribed and Inscribed Circles of a Triangle
A circumscribed circle passes through all three vertices of a triangle. To construct it, draw the perpendicular bisectors of any two sides of \triangle ABC. They meet at O. With O as centre and OA (or OB or OC) as radius, draw the circle. Point O is the circumcentre, and OA = OB = OC is the circumradius.
An inscribed circle touches all three sides of the triangle. Draw the bisectors of any two angles; they meet at I. From I, drop ID perpendicular to a side. With I as centre and ID as radius, draw the circle. Point I is the incentre, and ID is the in radius.
Each construction works because of a concurrence property. The perpendicular bisectors of the sides meet at one point that is equidistant from the vertices. The angle bisectors meet at one point that is equidistant from the sides. The method is the same whether the triangle is equilateral or any other shape.
Worked example. For a triangle with AB = 5 cm, AC = 4.5 cm and \angle BAC = 70^\circ, first draw the perpendicular bisectors of AB and AC. Their intersection O is the circumcentre. The circumscribing circle has centre O and radius OA. To inscribe a circle in the same triangle, bisect angles A and B to find I, drop ID perpendicular to BC, then draw the circle with centre I and radius ID.
The common mistake is swapping the two constructions. Perpendicular bisectors build the circumscribed circle. Angle bisectors build the inscribed circle. Never reverse them. This pair of constructions appears again in later chapters of the Class 10 ICSE Maths book.
19.3 Circumscribing and Inscribing a Circle on Regular Hexagon
Construction 5 draws the circumscribing circle of a regular hexagon. Draw the perpendicular bisectors of any two sides. Their intersection is the centre of the circle. With that centre and its distance from any vertex as radius, draw a circle that passes through all six vertices.
You need each interior angle before constructing the hexagon. For a regular polygon with n sides:
\left(\frac{2n - 4}{n}\right) \times 90^\circ = \left(\frac{2 \times 6 - 4}{6}\right) \times 90^\circ = 120^\circ
For n = 6, each interior angle is 120^\circ. The formula comes from dividing the polygon into triangles from one vertex.
Alternative method. When a circle circumscribes a regular hexagon, the radius always equals the side length. Draw any two main diagonals, such as AD and FC. They meet at the centre O. With O as centre and OA as radius, draw the circle.
Construction 6 draws the inscribing circle. Draw the bisectors of two interior angles; they meet at I. From I, draw IP perpendicular to a side. With I as centre and IP as radius, the circle touches every side.
Worked example. For a regular hexagon of side 5 cm, the interior angle formula gives 120^\circ. Since the circumradius equals the side, the circumscribing circle has radius 5 cm. For the inscribing circle, bisect two interior angles to locate I; the radius is the perpendicular from I to any side.
Edge case. If the hexagon is not regular, a circumscribing or inscribing circle is not always possible. These constructions assume a regular hexagon, where all sides and all angles are equal.
What to practise from Circles Class 10 ICSE
Exercise 19 has 22 questions. They group by construction type, so you can practise one skill at a time.
- Tangents from an exterior point: questions 1, 2 and 18.
- Two tangents with a given angle between them: questions 3 and 4.
- Circumscribed and inscribed circles of equilateral triangles: questions 5, 7, 13 and 14.
- Triangle constructions with a circumscribing or inscribed circle: questions 6, 8, 9, 19, 20, 21 and 22.
- Theory questions on the circumcentre and incentre: questions 10 and 11.
- An incentre chord question: question 12.
- Regular hexagon constructions: questions 15, 16 and 17.
Questions 17 to 22 are past board questions, marked 2010, 2011, 2012, 2013, 2014 and 2015. They show how the board frames each construction type. Practise them after the earlier questions.
Circles Class 10 ICSE frequently asked questions
How do you construct a tangent from an exterior point to a circle in circles class 10 ICSE?
Join the exterior point P to the centre O. Draw a circle with OP as its diameter; it cuts the given circle at A and B. Join PA and PB. Both are the required tangents. The length of each is PA = \sqrt{OP^2 - OA^2}.
What is the difference between the circumcentre and the incentre of a triangle?
The circumcentre is the meeting point of the perpendicular bisectors of the sides. It is equidistant from the vertices, so it is the centre of the circumscribed circle. The incentre is the meeting point of the angle bisectors. It is equidistant from the sides, so it is the centre of the inscribed circle.
Why is the radius of the circle circumscribing a regular hexagon equal to its side?
The chapter states this property directly: whenever a circle circumscribes a regular hexagon, its radius is always equal to the side of the hexagon. Because that holds, you can skip the perpendicular bisectors and find the centre by drawing two main diagonals, such as AD and FC, which meet at the centre O.
Which constructions questions from this chapter have appeared in the ICSE board exam?
Questions 17 to 22 of Exercise 19 carry board year labels from 2010 to 2015. They cover a regular hexagon with a circumscribed circle, tangents from an exterior point, triangles with circumscribed circles, an incircle construction, and a locus construction. Practise these last as a check on your construction skills.
Reference: ICSE Class 10 Mathematics textbook material, chapter 19: Constructions Circles — CISCE-aligned.
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