Circles

Circles is a crucial chapter in Class 10 Mathematics that deals with the properties and applications of circles.

Key Concepts:

Circle: A circle is a geometric figure consisting of all points in a plane that are equidistant from a fixed point called the center.
Diameter: A line segment passing through the center of a circle and connecting two points on its circumference..
Secant: A line that intersects the circumference of a circle at two distinct points.
Inscribed Angle: An angle formed by two chords intersecting on the circumference of a circle.
Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on the circumference of a circle.

Theorems and Properties:

Thales’ Theorem: A diameter subtends a right angle at any point on the circumference of a circle.
Tangent-Radius Property: The tangent to a circle is perpendicular to the radius drawn to the point of contact.
Intersecting Chords Theorem: If two chords of a circle intersect inside the circle, then the product of the segments of one chord is equal to the product of the segments of the other chord.
Tangent-Secant Theorem: If a tangent and a secant intersect outside the circle, then the square of the tangent is equal to the product of the external segment and the whole secant.

Applications:

Real-world problems involving circles: Finding the radius, diameter, circumference, or area of a circle.
Solving geometric problems related to tangents, secants, and chords.
Applications in engineering, architecture, and other fields.

In essence, the chapter on circles provides a comprehensive understanding of the properties and relationships associated with circles, equipping students with the necessary tools to solve various geometric problems.

Exercise 10.1

1. How many tangents can a circle have?

Ans : A circle can have infinitely many tangents.

2. Fill in the blanks:

(i) A tangent to a circle intersects it in ………… point(s).

(ii) A line intersecting a circle in two points is called a ………… .

(iii) A circle can have ………………. parallel tangents at the most.

(iv) The common point of a tangent to a circle and the circle is called ………

Ans :

(i) one

(ii) secant

(iii) two

(iv) point of contact

3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is

(a) 12 cm

(b) 13 cm

(c) 8.5 cm

(d) √119 cm

Ans :

Therefore, triangle OPQ is a right-angled triangle with ∠OPQ = 90°.

Using the Pythagorean theorem: PQ² = OQ² – OP²

Given, OP = 5 cm (radius of the circle) and OQ = 12 cm

Substituting the values: PQ² = 12² – 5² = 144 – 25 = 119

PQ = √119 cm

Therefore, the length of PQ is √119 cm.

The correct answer is (D) √119 cm.

4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Ans :

Exercise 10.2

1. In Q.1 to 3 choose the correct option and give justification.

1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

Ans :

The correct answer is (a) 7 cm.

Justification:

Let the center of the circle be O, and let the point of contact of the tangent be P. Since the tangent is perpendicular to the radius at the point of contact, triangle OPQ is a right-angled triangle with ∠OPQ = 90°.

Using the Pythagorean theorem:

OQ² = OP² + PQ²

25² = OP² + 24²
625 = OP² + 576
OP² = 49
OP = 7 cm

Therefore, the radius of the circle is 7 cm.

2. In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to

(a) 60°

(b) 70°

(c) 80°

(d) 90°

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q1

Ans :

The correct answer is (b) 70°.

Justification:

Property of tangents: The angle between a tangent and the radius drawn to the point of contact is 90°.
Let the points of contact of the tangents TP and TQ be P and Q, respectively.
Then, ∠OPQ = ∠OQT = 90° (where O is the center of the circle)
In quadrilateral OPQT, the sum of all angles is 360°.
Therefore, ∠PTQ + ∠POQ + ∠OPQ + ∠OQT = 360°
Substituting the known angles: ∠PTQ + 110° + 90° + 90° = 360°
∠PTQ + 290° = 360°
∠PTQ = 70°

Therefore, ∠PTQ is equal to 70°.

3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to

(a) 50°

(b) 60°

(c) 70°

(d) 80°

Ans :

The correct answer is (a) 50°.

Justification:

Property of tangents: The angle between a tangent and the radius drawn to the point of contact is 90°.
Let the points of contact of the tangents PA and PB be A and B, respectively.
Then, ∠OAP = ∠OBP = 90° (where O is the center of the circle)
In quadrilateral OAPB, the sum of all angles is 360°.
Therefore, ∠POA + ∠AOB + ∠OBP + ∠OPB = 360°
Substituting the known angles: ∠POA + 80° + 90° + 90° = 360°
∠POA + 260° = 360°
∠POA = 100°

However, we need to find the angle ∠POA, which is half of ∠AOB.

∠POA = (1/2) * ∠AOB
= (1/2) * 100° = 50°

Therefore, ∠POA is equal to 50°.

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

Ans :

Given:

A circle with center O.
AB is a diameter of the circle.
Tangents PA and PB are drawn at points A and B respectively.

To Prove:

PA || PB

Proof:

Property of tangents: The tangent to a circle is perpendicular to the radius drawn to the point of contact.
- ∠OAP = ∠OBP = 90° (where O is the center of the circle)
Angle in a semicircle:

∠AOB = 90°

Using the property of alternate interior angles:
- Here, ∠OAP = ∠AOB = 90° (proved above)
- These are alternate interior angles with respect to the transversal AB.
Therefore, PA || PB.

5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Ans :

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

Ans :

Since the tangent is perpendicular to the radius at the point of contact, triangle OPQ is a right-angled triangle with ∠OPQ = 90°.

Using the Pythagorean theorem:

OQ² = OP² + PQ²

Given, OQ = 5 cm and PQ = 4 cm. Substituting these values:

5² = OP² + 4²
25 = OP² + 16
OP² = 9
OP = 3 cm

7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Ans :

Let the larger circle be C1 with center O and radius 5 cm, and the smaller circle be C2 with the same center O and radius 3 cm.

Let the chord of the larger circle which touches the smaller circle be AB. Since AB is tangent to the smaller circle, it is perpendicular to the radius OC drawn to the point of contact C.

Therefore, triangle OAC is a right-angled triangle with ∠AOC = 90°.

Using the Pythagorean theorem:

OA² = OC² + AC²
5² = 3² + AC²
25 = 9 + AC²
AC² = 16
AC = 4 cm

Since AB = 2AC, the length of the chord AB is 2 * 4 = 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q8

Ans :

Given:

The circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

To Prove:

AB + CD = AD + BC

Proof:

Property of tangents: The length of tangents drawn from an external point to a circle are equal.

Therefore:

AP = AS
BP = BQ
CQ = CR
DR = DS

Now, adding these equations:

AP + BP + CQ + DR = AS + BQ + CR + DS
(AP + BP) + (CQ + DR) = (AS + DS) + (BQ + CR)
AB + CD = AD + BC

9. In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2 Q9

Ans :

Given:

AB is another tangent with the point of contact C, intersecting XY at A and X’Y’ at B.

To Prove:

∠AOB = 90°

Proof:

Property of tangents: The tangent to a circle is perpendicular to the radius drawn to the point of contact.
- ∠OAC = ∠OBC = 90°
Consider quadrilateral OACB:
- ∠OAC + ∠AOC + ∠COB + ∠OBC = 360° (sum of angles in a quadrilateral)
- 90° + ∠AOC + ∠COB + 90° = 360°
- ∠AOC + ∠COB = 180°
Since XY and X’Y’ are parallel, ∠AOC and ∠COB are alternate interior angles.
- Therefore, ∠AOC = ∠COB
Substituting ∠AOC = ∠COB in the equation ∠AOC + ∠COB = 180°:
- 2∠AOC = 180°
- ∠AOC = 90°

Hence, ∠AOB = ∠AOC + ∠COB = 90° + 90° = 180°

Therefore, ∠AOB = 90°.

10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Ans :

11. Prove that the parallelogram circumscribing a circle is a rhombus.

Ans :

Given:

A parallelogram ABCD circumscribes a circle.
The circle touches sides AB, BC, CD, and DA at points P, Q, R, and S respectively.

Proof:

Property of tangents: The length of tangents drawn from an external point to a circle are equal.

Therefore:

AP = AS
BP = BQ
CQ = CR
DR = DS

Now, since ABCD is a parallelogram, we know:

AB = CD and AD = BC

Substituting the values from the property of tangents:

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

Since AP = AS, BP = BQ, CQ = CR, and DR = DS, we can simplify the equations:

2AP = 2CR
AP = CR

Similarly, we can prove that:

BP = DS

Therefore, AP = BP = CR = DS.

Since all sides of the parallelogram ABCD are equal, it is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.