# NCERT Solutions for Class 10 Maths Chapter 10

NCERT Solutions for class 10 Maths Chapter 10 Circles all Exercises in English Medium and Hindi Medium or View in Video Format for CBSE, UP Board as well as MP Board students. These solutions are applicable for the session 2020-21 onward. Students can download 10th Maths Solutions in PDF format free to use offline. UP Board students are now using NCERT Books for their exams. So, they can download UP Board Solutions for Class 10 Maths Chapter 10 all exercise from this page of Tiwari academy. Download UP Board Solution and NCERT Solutions Apps 2020-21 based on latest NCERT Books for academic session 2020-2021.

## NCERT Solutions for class 10 Maths Chapter 10

 Class: 10 Maths (English and Hindi Medium) Chapter 10: Circles

### 10th Maths Chapter 10 Solutions

NCERT Solutions for class 10 Maths Chapter 10 Circles is given below to free download in PDF or use online without downloading. Download UP Board Solutions for Class 10 Maths which are updated for current academic session 2020-21 for all students who are using latest NCERT Books 2020-21.

• ### 10th Maths Chapter 10 Exercise 10.2

#### Class 10 Maths Exercise 10.1 Explanation and Solutions in Videos

Class 10 Maths Exercise 10.1 Explanation and Solution
Class 10 Maths Chapter 10 Exercise 10.1 Solution

#### Class 10 Maths Exercise 10.2 Explanation and Solutions in Videos

Class 10 Maths Exercise 10.2 Explanation and Solution
Class 10 Maths Chapter 10 Exercise 10.2 Solution

### NCERT Solutions for Class 10 Maths Chapter 10

• NCERT Solutions 10th MathsExercise 10.1Read more
• NCERT Solutions 10th MathsExercise 10.2Read more
##### What do you understand by a Circle?

Circle: A circle is the set of points in a plane which are at the same distance from a fixed point in the plane. The fixed point is called the centre of the circle and the line segment joining the centre & a point on circle is called radius.

##### What is tangent to a circle?

Tangent to a circle: If a line drawn in the plane of a circle intersects the circle in one and only one point, then the line is called a tangent to the circle and the point at which the line intersect the circle is called the point of contact of the line with the circle.

##### How many theorems are there in Class 10 Chapter 10 Circles to prove?

Theorems on tangent to a circle:
(i). A tangent to a circle is perpendicular to the radius drawn from point of contact or A line drawn perpendicular to a radius at its end point on the circle is a tangent to the circle.
(ii). the tangents drawn to a circle from a point in the exterior of the circle are equal in length or the lengths of tangents drawn from an external point to a circle are equal.

##### How many tangents to a circle are possible?

Number of tangents to a circle: There are the following cases:
1. There is no tangent to a circle passing through a point lying inside the circle.
2. There is one and only one tangent to a circle passing through a point lying on the circle.
3. There are exactly two tangents to a circle through a point lying outside the circle.

#### Previous Years Questions

ONE MARK QUESTIONS
1. Prove that tangents drawn at the ends of a diameter of a circle are parallel to each other. [CBSE 2017]
THREE MARKS QUESTIONS
1. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that angle PTQ = 2 (angle OPQ). [CBSE 2017]
FOUR MARKS QUESTIONS
1. Prove that the lengths of tangents drawn from an external point to a circle are equal. [CBSE 2017]

##### Historical Facts!
1. The circle is the most primitive and rudimentary of all human inventions. It is the corner stone in the foundation of science and technology. It is the basic tool used by greatest artists and architects in the history of mankind.
2. The ancient Indian symbol of a circle, with a dot in the middle, known as ‘BINDU’ was probably instrumental in the use of circle as a representation of the concept zero.
3. The ratio of the circumference of a circle to its diameter is always a constant. This constant is universally denoted by Greek letter π.

### Important Questions on Class 10 Maths Chapter 10

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
Let O be the centre of the circle.
Given: OQ = 25cm and PQ = 24 cm
We know that the radius is perpendicular to tangent. Therefore, OP ⊥ PQ
In ∆OPQ, By Pythagoras theorem,
OP^2 + PQ^2 = OQ^2
⇒ OP^2 + 24^2 = 25^2
⇒ OP^2 = 625 − 576
⇒ OP^2 = 49
⇒ OP = 7
Therefore, the radius of circle is 7 cm. Hence, the option (A) is correct.
एक वृत्त की कितनी स्पर्श रेखाएँ हो सकती हैं?
एक वृत्त की अनन्त स्पर्श रेखाएँ हो सकती हैं क्योंकि एक वृत्त में उस पर असीम बिंदुओं की संख्या होती है और हर बिंदु पर एक स्पर्शरेखा खींची जा सकती है।
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let AB is diameter, PQ and RS are tangents drawn at ends of diameter.
We know that the radius is perpendicular to tangent. Therefore, OA ⊥ RS and OB ⊥ PQ.
∠OAR = 90° and ∠OAS = 90°
∠OBP = 90° and ∠OBQ = 90°
From the above, we have
∠OAR = ∠OBQ [Alternate angles]
∠OAS = ∠OBP [Alternate angles]
Since, alternate angles are equal.
Hence, PQ is parallel to PS.
एक बिंदु A से, जो एक वृत्त के केंद्र से 5 cm दूरी पर है, वृत्त पर स्पर्श रेखा की लंबाई 4 cm है। वृत्त की त्रिज्या ज्ञात कीजिए।
माना O केंद्र वाले वृत्त पर AB एक स्पर्श रेखा है जो वृत्त को B पर स्पर्श करती है।
दिया है: OA = 5cm और AB = 4 cm
हम जानते हैं कि त्रिज्या, स्पर्श रेखा पर लम्ब होती है।
इसलिए, ∆ABO में, OB ⊥ AB
∆ABO में, पाइथागोरस प्रमेय से,
AB^2 + BO^2 = OA^2
4^2 + BO^2 = 5^2
16 + BO^2 = 25
BO^2 = 9
BO = 3
इसलिए, वृत्त की त्रिज्या 3 cm है।
सिद्ध कीजिए कि किसी बाह्य बिंदु से किसी वृत्त पर खींची गई स्पर्श रेखाओं के बीच का कोण स्पर्श बिंदुओं को मिलाने वाले रेखाखण्ड द्वारा केंद्र पर अंतरित कोण का संपूरक होता है।
माना O केंद्र वाले वृत्त पर PA तथा PB स्पर्श रेखाएँ हैं जो वृत्त को A और B पर स्पर्श करती हैं। OA और OB को मिलाया।
हम जानते हैं कि त्रिज्या, स्पर्श रेखा पर लम्ब होती है।
इसलिए, ∠OAP = 90° और ∠OBP = 90°
चतुर्भुज OAPB में,
∠OAP +∠APB+∠PBO +∠BOA = 360º
90º + ∠APB + 90º + ∠BOA = 360º
∠APB + ∠BOA = 180º
इस प्रकार यह सिद्ध हो जाता है कि किसी बाह्य बिंदु से किसी वृत्त पर खींची गई स्पर्श रेखाओं के बीच का कोण स्पर्श बिंदुओं को मिलाने वाले रेखाखण्ड द्वारा केंद्र पर अंतरित कोण का संपूरक होता है।       