Chapter – 10

Circles

In this post we have given the detailed notes of class 10 Maths Chapter 10 (Circles) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 10: Circles

Circle – Basic Terms

A circle is the locus (collection) of all points in a plane that are at a constant distance from a fixed point. The fixed point is called the centre of the circle, and the constant distance is called its radius.

• Chord: A line segment joining any two points on a circle. The diameter is the longest chord of a circle.
• Diameter: A chord passing through the centre of the circle; it is twice the radius (d = 2r).
• Arc: A part of the circumference of the circle.
• Circumference: The total length of the boundary of the circle (its perimeter).
• Segment: The region between a chord and the corresponding arc (major segment and minor segment).
• Sector: The region enclosed by two radii and the corresponding arc (major sector and minor sector).
• Secant: A line that intersects a circle at exactly two points.
• Tangent: A line that touches a circle at exactly one point.
• Semicircle: Half of a circle, cut off by a diameter.

Tangent to a Circle

A tangent to a circle is a line that intersects (touches) the circle at only one point. The point at which the tangent touches the circle is called the point of contact.

Note: There is one and only one tangent to a circle at a given point on the circle. A tangent can be thought of as a special (limiting) case of a secant, where the two end points of the corresponding chord coincide.

Number of Tangents from a Point to a Circle

Given a circle and a point, there are exactly three possibilities:

• Case I – Point inside the circle: No tangent can be drawn to a circle which passes through a point lying inside the circle. Every line through such a point is a secant.
• Case II – Point on the circle: There is exactly one tangent to a circle passing through a point lying on the circle.
• Case III – Point outside the circle: There are exactly two tangents to a circle through a point lying outside the circle.

The length of the segment of the tangent from the external point P to the point of contact is called the length of the tangent from the point P to the circle.

Theorem 10.1

Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent XY to the circle at a point P.

To prove: OP ⊥ XY.

Proof: Take a point Q, other than P, on the tangent XY, and join OQ. The point Q must lie outside the circle (because if Q were inside the circle, XY would become a secant of the circle and not a tangent to it). Therefore, OQ is longer than the radius OP of the circle, i.e.,

OQ > OP

Since this happens for every point Q on the line XY except P, OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY (the shortest distance from a point to a line is the perpendicular distance).

Note: This theorem also tells us that at any point on a circle there can be one and only one tangent. The line containing the radius through the point of contact is also called the ‘normal’ to the circle at that point.

Theorem 10.2

Statement: The lengths of the tangents drawn from an external point to a circle are equal.

Given: A circle with centre O, a point P lying outside the circle, and two tangents PQ and PR from P to the circle, where Q and R are the points of contact.

To prove: PQ = PR.

Proof: Join OP, OQ and OR. Then ∠OQP and ∠ORP are right angles (by Theorem 10.1, since a tangent is perpendicular to the radius through the point of contact). Now in the right triangles OQP and ORP:

OQ = OR  (radii of the same circle)

OP = OP  (common side)

∴ ΔOQP ≅ ΔORP  (by RHS congruence rule)

This gives PQ = PR  (by CPCT, i.e. corresponding parts of congruent triangles)

Remarks:

• This theorem can also be proved using the Pythagoras Theorem: PQ² = OP² − OQ² = OP² − OR² = PR² (since OQ = OR), which gives PQ = PR.
• Since ΔOQP ≅ ΔORP, we also get ∠OPQ = ∠OPR. This means OP bisects the angle between the two tangents, i.e., the centre of the circle lies on the bisector of the angle between the two tangents drawn from an external point.

Worked Examples

Example 1

Question: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution: Let AB and CD be two tangents drawn at the ends of a diameter XY of a circle with centre O, touching the circle at X and Y respectively.

Since a tangent at a point on a circle is perpendicular to the radius through that point:

OX ⊥ AB  ⇒  ∠OXA = 90°  …(i)

OY ⊥ CD  ⇒  ∠OYC = 90°  …(ii)

Adding (i) and (ii): ∠OXA + ∠OYC = 90° + 90° = 180°

Since XY is a straight line (the diameter) intersecting AB and CD, and the co-interior (allied) angles on the same side of the transversal add up to 180°, AB and CD must be parallel.

∴ AB ∥ CD  (Proved)

Example 2

Question: A point A is at a distance of 5 cm from the centre of a circle. The length of the tangent drawn from A to the circle is 4 cm. Find the radius of the circle.

Solution: Distance of point A from centre, OA = 5 cm

Length of tangent AB = 4 cm

Radius of the circle, OB = ?

Since the tangent is perpendicular to the radius at the point of contact, triangle OAB is right-angled at B. By the Pythagoras Theorem:

OA² = OB² + AB²

5² = OB² + 4²

25 − 16 = OB²

OB² = 9

OB = 3 cm

∴ The radius of the circle is 3 cm.

Example 3

Question: The radii of two concentric circles are 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution: Let O be the common centre. AB is a chord of the larger circle that touches the smaller circle at point M.

Radius of larger circle, OA = 5 cm

Radius of smaller circle, OM = 3 cm

Since AB touches the smaller circle at M, OM ⊥ AB (the radius to the point of contact is perpendicular to the tangent chord).

In right triangle OMA, by the Pythagoras Theorem:

OA² = OM² + AM²

5² = 3² + AM²

25 − 9 = AM²

AM² = 16  ⇒  AM = 4 cm

Since the perpendicular from the centre bisects the chord, AB = 2 × AM = 2 × 4 = 8 cm

∴ The length of the chord is 8 cm.

Example 4

Question: A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Solution: Let the circle with centre O touch the sides AB, BC, CD and DA of the quadrilateral at points P, Q, R and S respectively.

Since the lengths of tangents drawn from an external point to a circle are equal (Theorem 10.2):

AP = AS  …(i)  (tangents from A)

BP = BQ  …(ii)  (tangents from B)

CR = CQ  …(iii)  (tangents from C)

DR = DS  …(iv)  (tangents from D)

Adding equations (i), (ii), (iii) and (iv):

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

∴ AB + CD = AD + BC  (Proved)

Example 5

Question: Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠AOB + ∠APB = 180°.

Solution: Since the tangent at any point of a circle is perpendicular to the radius through the point of contact:

OA ⊥ AP  ⇒  ∠OAP = 90°  …(i)

OB ⊥ BP  ⇒  ∠OBP = 90°  …(ii)

Now, OAPB is a quadrilateral, so the sum of its interior angles is 360°:

∠OAP + ∠AOB + ∠OBP + ∠APB = 360°

90° + ∠AOB + 90° + ∠APB = 360°   [using (i) and (ii)]

∠AOB + ∠APB = 360° − 180° = 180°

∴ ∠AOB + ∠APB = 180°  (Proved)

Key Points to Remember

• A tangent to a circle touches it at exactly one point, called the point of contact; a secant touches it at two points.
• There is no tangent to a circle through a point lying inside it.
• There is exactly one tangent to a circle through a point lying on the circle.
• There are exactly two tangents to a circle through a point lying outside it, and their lengths are equal.
• Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.
• The centre of a circle lies on the bisector of the angle between two tangents drawn from an external point.
• A tangent is a special case of a secant where both points of intersection coincide.

We hope that class 10 Maths Chapter 10 (Circles) notes in English helped you. If you have any query about class 10 Maths Chapter 10 (Circles) notes in English or about any other notes of class 10 Maths in English, so you can comment below. We will reach you as soon as possible…