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11. Conic Sections Mathematics class 11 in English Medium ncert book solutions Exercise 11.1
11. Conic Sections Exercise 11.1 – Complete NCERT Book Solutions for Class 11 Mathematics (English Medium). Get all chapter explanations, extra questions, solved examples and additional practice questions for 11. Conic Sections Exercise 11.1 to help you master concepts and score higher.
11. Conic Sections Mathematics class 11 in English Medium ncert book solutions Exercise 11.1
NCERT Solutions for Class 11 Mathematics play an important role in helping students understand the concepts of the chapter 11. Conic Sections clearly. This chapter includes the topic Exercise 11.1, which is essential from both academic and examination point of view. The solutions provided here are prepared strictly according to the latest NCERT syllabus and follow the guidelines of CBSE to ensure accuracy and relevance. Each question is explained in a simple and student-friendly manner so that learners can grasp the concepts without confusion. These NCERT Solutions are useful for regular study, homework help, and exam preparation. All textbook questions are solved step by step to improve problem-solving skills and conceptual clarity. Students of Class 11 studying Mathematics can use these solutions to revise important topics, understand difficult questions, and practise effectively before examinations. The chapter 11. Conic Sections is explained in a structured way, making it easier for students to connect the theory with the topic Exercise 11.1. By studying these updated NCERT Solutions for Class 11 Mathematics, students can build a strong foundation, boost their confidence, and score better marks in school and board exams.
11. Conic Sections
Exercise 11.1
Exercise 11.1 (Conic Sections)
Q1. Find the equation of the circle with centre (0, 2) and radius 2.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0
Q2. Find the equation of the circle with centre (–2, 3) and radius 4.
Solution:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2
⟹ x2 + 4x + 4 + y2 – 6y + 9 = 16
⟹ x2 + y2 + 4x – 6y – 3 = 0
Q3.
Q10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 …………………. (1)
(6 – h)2 + (5 – k)2 = r2 …………………. (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
∴ 4h + k = 16 …………………………………… (3)
From equations (1) and (2), we obtain
(4 – h)2 + (1 – k)2 = (6 – h)2 + (5 – k)2
⇒ 16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 ………………………………… (4)
On solving equations (3) and (4), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (1), we obtain
(4 – 3)2 + (1 – 4)2 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
⇒ 𝑟=√10
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = (√10)2
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 + y2 – 6x – 8y + 15 = 0
The required equation for given circle is x2 + y2 – 6x – 8y + 15 = 0
See other sub-topics of this chapter:
1. Exercise 11.1 2. Exercise 11.2 3. Exercise 11.3 4. Exercise 11.4 5. Miscellaneous Exercise on Chapter - 11