Circles

Important Questions

Here are some important questions for Class 10 Mathematics Chapter 10, Circles, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2024-25. These questions cover various types of problems and are designed to assist students in understanding Circles better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Circles.

Table of Contents

In Chapter 10 of Class 10 Mathematics, Circles, In this chapter, we explore the properties of tangents to a circle at the point of contact. Two important properties are:

- (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
- (Prove) The lengths of tangents drawn from an external point to a circle are equal.

A tangent to a circle is a straight line that touches the circle at exactly one point, known as the point of tangency. This point of contact is where the tangent line intersects the circle’s circumference. The tangent line is always perpendicular to the radius drawn to the point of tangency.

**Ans.**(b) 9 cm

**Explanation:**

Since, PA and PB are tangents to the circle from point P.

Therefore PA = PB …(i)

Also, CD is the tangent touching the circle at Q.

Therefore, CA = CQ and DQ = DB …(ii)

Since, tangents from external points are equal.

Now, PA = PB = 12 cm

CA = CQ = 3 cm

PC = PA - CA

= 12 - 3 = 9 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

**Ans.** (b) 2 cm

**Explanation:**

An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle.

OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA.

OPBQ is a square. ( ∵ ∠B - 90^o)\\
Let r be the radius of the circle

PB = BQ = r

AR = AP = 8 - r,

CQ = CR = 6 - r

AC = AR + CR

⇒ 10 = 8 - r + 6 - r

⇒ 10 = 14 - 2r

⇒ 2r = 14 - 10 = 4

⇒ r = 2

**Ans.**PR = 19 cm**Explanation:**

Let the line QR intersect the bigger circle at S.

Join PS.

O is the mid-point of PQ. [∵ PQ is a diameter of the bigger circle]QR is a tangent to the smaller circle and OR is a radius through the point of contact R.

∴ OR ⊥ QR ⇒ OR ⊥ QS

Since, OR is perpendicular to a chord QS of the bigger circle.

∴ QR = RS

[ ∵ Perpendicular from the centre to a chord bisects the chord]

⇒ Risthemid - point of Qs.

∴ In △QSP, O is the mid-point of PQ and R is the mid-point of QS. \\\text{OR} = \dfrac{1}{2} \text{Ps} \\[∵ segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ PS = 2OR = 2 × 8 cm = 16 cm

\\
\text{In right △OQR, OR}^2 + QR^2 = OQ^2\\[4.5 bp]⇒ 8^2 + QR^2 = 13^2\\[4.5 bp]⇒ 64 + QR^2 = 169\\[4.5 bp]⇒ QR^2 = 169 – 64 = 105 \\[4.5 bp]QR = \sqrt{105} \\[4.5 bp]\therefore \space RS = QR = \sqrt{105}\\[4.5 bp]\text{In △PRS, PR}^2 = \text{RS}^2\text{ + PS}\\[4.5 bp]= \sqrt{105}^2 + 16^2\\[4.5 bp]⇒ PR = 19 cm.

**Ans.** Perimeter of △PMN = 10 cm.

**Explanation:**

Here, MT = MQ [Tangents from point M] ...(i)

And NT = NR [Tangents from point N] ...(ii)

Now, PQ + PR = PM + MQ + PN + NR

= PM + MT + PN + NT [Using eq. (i) and (ii)]
= PM + PN + (MT + NT)

= PM + PN + MN = Perimeter of DPMN

∴ Perimeter of △PMN = 5 + 5

= 10 cm.

**Ans.** Perimeter of △PCD = 28 cm

**Explanation:**

Given, PA and PB are the tangents drawn from a point P outside the circle with centre O.CD is another tangents to the circle at point E which intersects PA and PB at C and D respectively.

PA and PB are the tangents to the circle from P

So, PA = PB = 14 cm

Now, CA and CE are the tangents from C to the circle.

∴ CA = CE …(i)

Now, perimeter of △PCD

= PC + PD + CD

= PC + PD + CE + DE

= PC + CE + PD + DE

= PC + CA + PD + DB

{From (i) and (ii)}

= PA + PB

= 14 + 14

= 28 cm

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

If you want to get better at understanding the Circles chapter, you should check out oswal.io. This website offers lots of practice questions made to help you learn more effectively. By practicing these questions, you can improve how well you understand circles and get better at solving problems related to them. It’s a really helpful way to practice and feel more confident when dealing with circle concepts.

Ans: The diameter of a circle is twice the length of its radius. In other words, diameter = 2 × radius.

Ans: The circumference of a circle (C) is given by the formula C = 2πr, where ‘r’ is the radius. Also, the diameter (d) is twice the radius, so d = 2r.

Ans: Angle between a tangent and a radius drawn to the point of tangency is 90 degrees, making them perpendicular to each other.

Ans: In a circle, tangents drawn from an external point are equal in length. They are both perpendicular to the radius drawn to the point of contact on the circle.

Ans: The angle sum of a quadrilateral formed by four tangents drawn to a circle is 360 degrees.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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