NCERT Solutions Class 11 Maths Chapter 11 Conic Sections – Here are all the NCERT solutions for Class 11 Maths Chapter 11. This solution contains questions, answers, images, explanations of the complete chapter 11 titled Of Conic Sections taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 11 Conic Sections After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections in one place.
NCERT Solutions Class 11 Maths Chapter 11 Conic Sections
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 11 Conic Sections , Maths, Class 11.
Class | 11 |
Subject | Maths |
Book | Mathematics |
Chapter Number | 11 |
Chapter Name |
Conic Sections |
NCERT Solutions Class 11 Maths chapter 11 Conic Sections
Class 11, Maths chapter 11, Conic Sections solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Centre (0,2) and radius 2
Ans : The equation of a circle with centre (h, k) and radius r is given as \((x-h)^{2}+(y-k)^{2}=r^{2}\) It is given that centre (h, k) = (0, 2) and radius (r) = 2. Therefore, the equation of the circle is \(\begin{array}{l}{(x-0)^{2}+(y-2)^{2}=2^{2}} \\ {x^{2}+y^{2}+4-4 y=4} \\ {x^{2}+y^{2}-4 y=0}\end{array}\)
Q.2: Find the equation of the circle with centre (–2,3) and radius 4
Ans : The equation of a circle with centre (h, k) and radius r is given as \((x-h)^{2}+(y-k)^{2}=r^{2}\) It is given that centre (h, k) = (–2, 3) and radius (r) = 4. Therefore, the equation of the circle is \(\begin{array}{l}{(x+2)^{2}+(y-3)^{2}=(4)^{2}} \\ {x^{2}+4 x+4+y^{2}-6 y+9=16} \\ {x^{2}+y^{2}+4 x-6 y-3=0}\end{array}\)
Q.3: Find the equation of the circle with centre \(\left(\frac{1}{2}, \frac{1}{4}\right) \text { and radius } \frac{1}{12}\)
Ans : The equation of a circle with centre (h, k) and radius r is given as \((x-h)^{2}+(y-k)^{2}=r^{2}\) It is given that centre \((\mathrm{h}, \mathrm{k})=\left(\frac{1}{2}, \frac{1}{4}\right){\text { and radius }(\mathrm{r})}=\frac{1}{12}\) Therefore, the equation of the circle is \(\begin{array}{l}{\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{4}\right)^{2}=\left(\frac{1}{12}\right)^{2}} \\ {x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}=\frac{1}{144}} \\ {x^{2}-x+\frac{1}{4}+y^{2}-\frac{y}{2}+\frac{1}{16}-\frac{1}{144}=0} \\ {144 x^{2}-144 x+36+144 y^{2}-72 y+9-1=0} \\ {144 x^{2}-36 x+144 y^{2}-18 y+41=0} \\ {36 x^{2}-36 x+36 y^{2}-18 y+11=0} \\ {36 x^{2}+36 y^{2}-36 x-18 y+11=0}\end{array}\)
Q.4: Find the equation of the circle with centre (1,1) and radius 2
Ans : The equation of a circle with centre (h, k) and radius r is given as \((x-h)^{2}+(y-k)^{2}=r^{2}\) It is given that centre (h, k) = (1, 1) and radius (r) = \(\sqrt{2}\) Therefore, the equation of the circle is \(\begin{array}{l}{(x-1)^{2}+(y-1)^{2}=(\sqrt{2})^{2}} \\ {x^{2}-2 x+1+y^{2}-2 y+1=2} \\ {x^{2}+y^{2}-2 x-2 y=0}\end{array}\)
Q.5: Find the equation of the circle with centre (–a, –b) and radius \(\sqrt{a^{2}-b^{2}}\)
Ans : The equation of a circle with centre (h, k) and radius r is given as \((x-h)^{2}+(y-k)^{2}=r^{2}\) It is given that centre (h, k) = (–a, –b) and radius (r) = \(\sqrt{a^{2}-b^{2}}\). Therefore, the equation of the circle is \(\begin{array}{l}{(x+a)^{2}+(y+b)^{2}=\left(\sqrt{a^{2}-b^{2}}\right)^{2}} \\ {x^{2}+2 a x+a^{2}+y^{2}+2 b y+b^{2}=a^{2}-b^{2}} \\ {x^{2}+y^{2}+2 a x+2 b y+2 b^{2}=0}\end{array}\)
NCERT / CBSE Book for Class 11 Maths
You can download the NCERT Book for Class 11 Maths in PDF format for free. Otherwise you can also buy it easily online.
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