# NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.1) Exercise 11.1

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## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Ex 11.1) Exercise 11.1

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### Access NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections

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### Exercise 11.1

Exercise 11.1 Maths Class 11 includes the Introduction of the Chapter, Sections of a Cone, Circle, Ellipse, Parabola, Hyperbola, Degenerated Conic Sections, and Concepts Related to Circle. The NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 provided by Extramarks is a very useful resource for quick revisions. If the solutions are not explained in proper steps, students can have difficulty and doubts in the concepts and calculations of the chapter. Furthermore, Mathematics is a subject where students can score marks on the basis of step-wise marking. Therefore, Extramarks provides students with detailed step-by-step NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1.

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### NCERT Solutions for Class 11 Maths Chapters

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### NCERT Solution Class 11 Maths of Chapter 11 All Exercises

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### Important Topics Covered in Exercise 11.1 of Class 11 Maths NCERT Solutions

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Q.1 In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2
2. centre (–2,3) and radius 4
3.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\right)}\mathbf{\text{}}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\frac{\mathbf{1}}{\mathbf{12}}$

4.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{\right)}\mathbf{\text{}}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{\mathbf{2}}$

5.

$\mathbf{centre}\mathbf{\text{\hspace{0.17em}}}\mathbf{\left(}\mathbf{-}\mathbf{a}\mathbf{,}\mathbf{-}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{radius}\mathbf{}\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}$

Ans.
1.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(0,2\right)\text{and radius 2 is:}\\ {\left(\mathrm{x}-0\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={2}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{y}+4=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

2.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(-2,3\right)\text{and radius 4 is:}\\ {\left(\mathrm{x}+2\right)}^{2}+{\left(\mathrm{y}-3\right)}^{2}={4}^{2}\\ ⇒{\mathrm{x}}^{2}+4\mathrm{x}+4+{\mathrm{y}}^{2}-6\mathrm{y}+9=16\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+4\mathrm{x}-6\mathrm{y}-3=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

3.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(\frac{1}{2},\frac{1}{4}\right)\text{and radius}\frac{1}{12}\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\frac{1}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{1}{4}\right)}^{2}={\left(\frac{1}{12}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+\frac{1}{4}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{1}{16}=\frac{1}{144}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-\mathrm{x}+{\mathrm{y}}^{2}-\frac{1}{2}\mathrm{y}+\frac{11}{36}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}36{\mathrm{x}}^{2}+36{\mathrm{y}}^{2}-36\mathrm{x}-18\mathrm{y}+11=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

4.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(1,1\right)\text{and radius}\sqrt{\text{2}}\text{is:}\end{array}$ $\begin{array}{}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}{\left(\mathrm{x}-1\right)}^{2}+{\left(\mathrm{y}-1\right)}^{2}={\left(\sqrt{\text{2}}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-2\mathrm{x}+1+{\mathrm{y}}^{2}-2\mathrm{y}+1=2\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-2\mathrm{x}-2\mathrm{y}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

5.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ \mathrm{Equation}\text{of circle with centre}\left(-\mathrm{a},-\mathrm{b}\right)\text{and radius}\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^{2}}\text{is:}\\ \text{\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}+\mathrm{a}\right)}^{2}+{\left(\mathrm{y}+\mathrm{b}\right)}^{2}={\left(\sqrt{{\text{a}}^{\text{2}}-{\mathrm{b}}^{2}}\right)}^{2}\\ {\mathrm{x}}^{2}+2\mathrm{ax}+{\mathrm{a}}^{2}+{\mathrm{y}}^{2}+2\mathrm{by}+{\mathrm{b}}^{2}={\text{a}}^{\text{2}}-{\mathrm{b}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}+2\mathrm{ax}+2\mathrm{by}+2{\mathrm{b}}^{2}=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.2 In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2 + (y – 3)2 = 36
7. x2 + y2 – 4x – 8y – 45 = 0
8. x2 + y2 – 8x + 10y – 12 = 0
9. 2x2 + 2y2 – x = 0

Ans.

6.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{(x+5)}}^{\text{2}}\text{+(y}-{\text{3)}}^{\text{2}}\text{=36}...\left(\text{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=-5,\mathrm{k}=3{\text{and r}}^{\text{2}}=\text{36}\end{array}$

$\begin{array}{l}⇒\mathrm{h}=-5,\mathrm{k}=3\text{and r}=\text{6}\\ \text{Therefore, centre is}\left(-5,3\right)\text{and radius is 6.}\end{array}$

7.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{x}-8\mathrm{y}-45=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-4\mathrm{x}\right)+\left({\mathrm{y}}^{2}-8\mathrm{y}\right)-45=0\\ ⇒\left({\mathrm{x}}^{2}-4\mathrm{x}+{2}^{2}-{2}^{2}\right)+\left({\mathrm{y}}^{2}-8\mathrm{y}+{4}^{2}-{4}^{2}\right)-45=0\\ ⇒\left({\mathrm{x}}^{2}-4\mathrm{x}+{2}^{2}\right)-4+\left({\mathrm{y}}^{2}-8\mathrm{y}+{4}^{2}\right)-16-45=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-2\right)}^{2}+{\left({\mathrm{y}}^{2}-4\right)}^{2}-65=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-2\right)}^{2}+{\left({\mathrm{y}}^{2}-4\right)}^{2}={\left(\sqrt{65}\right)}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=2,\mathrm{k}=4\text{and r}=\sqrt{65}\\ \text{Therefore, centre is}\left(2,4\right)\text{and radius is}\sqrt{\text{65}}\text{.}\end{array}$

8.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-8\mathrm{x}+10\mathrm{y}-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}\right)+\left({\mathrm{y}}^{2}+10\mathrm{y}\right)-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}+{4}^{2}-{4}^{2}\right)+\left({\mathrm{y}}^{2}+10\mathrm{y}+{5}^{2}-{5}^{2}\right)-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-8\mathrm{x}+{4}^{2}\right)-16+\left({\mathrm{y}}^{2}+10\mathrm{y}+{5}^{2}\right)-25-12=0\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-4\right)}^{2}+{\left({\mathrm{y}}^{2}+5\right)}^{2}-53=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left({\mathrm{x}}^{2}-4\right)}^{2}+{\left({\mathrm{y}}^{2}+5\right)}^{2}={\left(\sqrt{53}\right)}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=4,\mathrm{k}=-5\text{and r}=\sqrt{53}\\ \text{Therefore, centre is}\left(4,-5\right)\text{and radius is}\sqrt{\text{53}}\text{.}\end{array}$

9.

$\begin{array}{l}\mathrm{The}\text{given equation of circle is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}2x}}^{\text{2}}+{\text{2y}}^{\text{2}}-\text{x}=0\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}-\frac{1}{\text{2}}\text{x}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}\right)+{\mathrm{y}}^{2}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}+\frac{1}{{4}^{2}}-\frac{1}{{4}^{2}}\right)+{\mathrm{y}}^{2}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{x}}^{2}-\frac{1}{2}\mathrm{x}+\frac{1}{{4}^{2}}\right)+{\mathrm{y}}^{2}=\frac{1}{{2}^{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\frac{1}{4}\right)}^{2}+{\mathrm{y}}^{2}=\frac{1}{{4}^{2}}...\left(\mathrm{i}\right)\\ \mathrm{Comparing}\text{equation}\left(\mathrm{i}\right)\text{with the equation of circle}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\text{and we get}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}=\frac{1}{4},\mathrm{k}=0\text{and r}=\frac{1}{4}\end{array}$

$\text{Therefore, centre is}\left(\frac{1}{4},0\right)\text{and radius is}\frac{1}{4}\text{.}$

Q.3 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.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{is passing through}\left(4,1\right)\text{and}\left(6,5\right)\text{then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}{\left(6-\mathrm{h}\right)}^{2}+{\left(5-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}​and equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\left(6-\mathrm{h}\right)}^{2}+{\left(5-\mathrm{k}\right)}^{2}\\ ⇒16-8\mathrm{h}+{\mathrm{h}}^{2}+1-2\mathrm{k}+{\mathrm{k}}^{2}=36-12\mathrm{h}+{\mathrm{h}}^{2}+25-10\mathrm{k}+{\mathrm{k}}^{2}\\ ⇒4\mathrm{h}+8\mathrm{k}=44\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}+2\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{​ centre}\left(\mathrm{h},\mathrm{k}\right)\text{lies on the line 4x}+\mathrm{y}=16.\\ \mathrm{So},4\mathrm{h}+\mathrm{k}=16...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\text{3 and k}=\text{4}\\ \text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(4-3\right)}^{2}+{\left(1-4\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{2}=1+9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=10\end{array}$

$\begin{array}{l}\mathrm{Then},\text{equation of circle is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-3\right)}^{2}+{\left(\mathrm{y}-4\right)}^{2}=10\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-6\mathrm{x}+9+{\mathrm{y}}^{2}-8\mathrm{y}+16=10\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-6\mathrm{x}-8\mathrm{y}+15=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.4 Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{is passing through}\left(2,3\right)\text{and}\left(-1,1\right)\text{then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{h}\right)}^{2}+{\left(3-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\\ \mathrm{and}\text{\hspace{0.17em}}{\left(-1-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{\hspace{0.17em}​and equation}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{h}\right)}^{2}+{\left(3-\mathrm{k}\right)}^{2}={\left(-1-\mathrm{h}\right)}^{2}+{\left(1-\mathrm{k}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4-4\text{\hspace{0.17em}}\mathrm{h}+{\mathrm{h}}^{2}+9-6\mathrm{k}+{\mathrm{k}}^{2}=1+2\mathrm{h}+{\mathrm{h}}^{2}+1-2\mathrm{k}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}}6\mathrm{h}-4\mathrm{k}=-11\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{h}+4\mathrm{k}=11...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{​ centre}\left(\mathrm{h},\mathrm{k}\right)\text{lies on the line x}-3\mathrm{y}=11.\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{h}-3\mathrm{k}=11...\left(\mathrm{v}\right)\\ \mathrm{Solving}\text{equation}\left(\mathrm{iv}\right)\text{and equation}\left(\mathrm{v}\right),\text{we get}\\ \text{h}=\frac{7}{2}\text{and k}=-\frac{\text{5}}{2}\end{array}$

$\begin{array}{l}\text{Putting value of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ {\left(2-\frac{7}{2}\right)}^{2}+{\left(3+\frac{\text{5}}{2}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{2}=\frac{9}{4}+\frac{121}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{130}{4}\\ \mathrm{Then},\text{equation of circle is:}\\ {\left(\mathrm{x}-\frac{7}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{\text{5}}{2}\right)}^{2}=\frac{130}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-7\mathrm{x}+\frac{49}{4}+{\mathrm{y}}^{2}+5\mathrm{y}+\frac{25}{4}=\frac{130}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{x}}^{2}+4{\mathrm{y}}^{2}-28\mathrm{x}+20\mathrm{y}=56\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-7\mathrm{x}+5\mathrm{y}-14=0\\ \mathrm{Which}\text{is the required equation of the circle.}\end{array}$

Q.5 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{a},0\right)\text{and radius 5 is:}\\ {\left(\mathrm{x}-\mathrm{a}\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\mathrm{x}-\mathrm{a}\right)}^{2}+{\mathrm{y}}^{2}={5}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Since},\text{circle}\left(\mathrm{i}\right)\text{passes through the point}\left(2,3\right).\text{\hspace{0.17em}}\mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(2-\mathrm{a}\right)}^{2}+{3}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4-4\mathrm{a}+{\mathrm{a}}^{2}+9=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-4\mathrm{a}=25-13\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-4\mathrm{a}-12=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{a}-6\right)\left(\mathrm{a}+2\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=6,-2\\ \mathrm{Case}\text{\hspace{0.17em}}1:\mathrm{When}\text{centre of circle is}\left(6,0\right):\\ \mathrm{The}\text{equation of circle is:-}\\ {\left(\mathrm{x}-6\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-12\mathrm{x}+36+{\mathrm{y}}^{2}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-12\mathrm{x}+{\mathrm{y}}^{2}+11=0\\ \mathrm{Case}\text{\hspace{0.17em}}2:\mathrm{When}\text{centre of circle is}\left(-2,0\right):\\ \mathrm{The}\text{equation of circle is:-}\\ {\left(\mathrm{x}+2\right)}^{2}+{\left(\mathrm{y}-0\right)}^{2}={5}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+4\mathrm{x}+4+{\mathrm{y}}^{2}=25\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+4\mathrm{x}+{\mathrm{y}}^{2}-21=0\\ \mathrm{Thus},\text{the equation of circle is}\\ {\mathrm{x}}^{2}+{\mathrm{y}}^{2}+4\mathrm{x}-21=0\text{​ or}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-12\mathrm{x}+11=0.\end{array}$

Q.6 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(\mathrm{h},\mathrm{k}\right)\text{and radius r is:}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^{2}+{\left(\mathrm{y}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the points}\left(0,0\right),\left(\mathrm{a},0\right)\text{and}\left(0,\mathrm{b}\right),\text{so}\\ {\left(0-\mathrm{h}\right)}^{2}+{\left(0-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}{}^{}\end{array}$

$\begin{array}{l}{\left(\mathrm{a}-\mathrm{h}\right)}^{2}+{\left(0-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-2\mathrm{ah}+{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iii}\right)\\ {\left(0-\mathrm{h}\right)}^{2}+{\left(\mathrm{b}-\mathrm{k}\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{b}}^{2}-2\mathrm{bk}+{\mathrm{k}}^{2}={\mathrm{r}}^{2}...\left(\mathrm{iv}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{a}}^{2}-2\mathrm{ah}+{\mathrm{h}}^{2}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{2}-2\mathrm{ah}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}\left(\mathrm{a}-2\mathrm{h}\right)=0\\ ⇒\text{\hspace{0.17em}}\mathrm{a}-2\mathrm{h}=0,\mathrm{a}\ne 0\\ ⇒\text{\hspace{0.17em}}\mathrm{h}=\frac{\mathrm{a}}{2}\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iv}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{h}}^{2}+{\mathrm{k}}^{2}={\mathrm{h}}^{2}+{\mathrm{b}}^{2}-2\mathrm{bk}+{\mathrm{k}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}-2\mathrm{bk}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}\left(\mathrm{b}-2\mathrm{k}\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}-2\mathrm{k}=0,\mathrm{b}\ne 0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{k}=\frac{\mathrm{b}}{2}\\ \mathrm{Putting}\text{​ values of h and k in equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\frac{\mathrm{b}}{2}\right)}^{2}={\mathrm{r}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}={\mathrm{r}}^{2}\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right):\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}{\left(\mathrm{x}-\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\mathrm{y}-\frac{\mathrm{b}}{2}\right)}^{2}=\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}\\ ⇒{\mathrm{x}}^{2}-\mathrm{ax}+\frac{{\mathrm{a}}^{2}}{4}+{\mathrm{y}}^{2}-\mathrm{by}+\frac{{\mathrm{b}}^{2}}{4}=\frac{{\mathrm{a}}^{2}}{4}+\frac{{\mathrm{b}}^{2}}{4}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-\mathrm{ax}-\mathrm{by}=0\\ \mathrm{This}\text{is the required equation of the circle.}\end{array}$

Q.7 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans.

$\begin{array}{l}\mathrm{Equation}\text{of circle with centre}\left(2,2\right)\text{and radius r is:}\\ {\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}={\mathrm{r}}^{2}...\left(\mathrm{i}\right)\\ \mathrm{Circle}\text{passes through the point}\left(4,5\right),\text{so}\\ {\left(4-2\right)}^{2}+{\left(5-2\right)}^{2}={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4+9={\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}13={\mathrm{r}}^{2}\\ \mathrm{Putting}{\text{value of r}}^{\text{2}}\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ {\left(\mathrm{x}-2\right)}^{2}+{\left(\mathrm{y}-2\right)}^{2}=13\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-4\mathrm{x}+4+{\mathrm{y}}^{2}-4\mathrm{y}+4=13\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}+{\mathrm{y}}^{2}-4\mathrm{x}-4\mathrm{y}=5\\ \mathrm{This}\text{is the equation of required circle.}\end{array}$

Q.8 Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Ans.

$\begin{array}{l}\mathrm{The}\text{equation of given circle is:}\\ {\text{x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{25}\end{array}$

$\begin{array}{l}\text{Putting x}=\text{2}.\text{5},\text{y}=\text{3}.\text{5 in L.H.S. of given equation, we get}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}}^{\text{2}}+{\text{y}}^{\text{2}}=\text{\hspace{0.17em}}{\left(\text{2}.\text{5}\right)}^{\text{2}}+{\left(\text{3}.\text{5}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6.25+12.25\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=18.50<25=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{So},\text{point}\left(–\text{2}.\text{5},\text{3}.\text{5}\right)\text{lies inside of the circle.}\\ \text{As distance of the point from centre is less than radius}\\ \text{of the circle.}\end{array}$

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### 1. Is it essential to practice every question of the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1?

Mathematics is a subject that requires ample amount of practice. Every question of the exercise contain different concepts and calculations. In order to score well in Mathematics Class 11 examinations and having a clarity of the concepts is very important. Also, practising these solutions enhance the mathematical skills and the calculation speed and accuracy of the students. Therefore, it is essential for them to practice each and every question of the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1.

### 2. Are the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 difficult?

No, the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 are not difficult. However, they help students understand the concepts of the chapter with a great ease as they are explained in proper steps. These solutions walk students through each step of the problem and help them understand the logic behind it. Practising these solutions thoroughly will help the students to gain higher marks in the Class 11 Mathematics examinations.

### 3. Is the NCERT Textbook enough to prepare the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1?

Yes, the NCERT Textbook is enough to build strong basic concepts of the students, so that they can solve any problem related to these concepts. Students should also practice sample papers and the past years’ question papers for a better preparation of the subject. Practising the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 help students to have a better hold of the concepts and eventually leads to better scores.

### 4. Are the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 available on Extramarks?

Yes, the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 are available on the Extramarks website. They are easily accessible as they are in the PDF format, and can also be downloaded on a variety of devices.

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Students can refer to the NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 provided by the Extramarks website to resolve their queries. Furthermore, students can subscribe to the Extramarks website and have access to learning modules such as Live Doubt-Solving Classes, In-Depth Performance Reports, Gamified Learning and much more. Students can also record their lectures for further assistance, which are very helpful in solving their doubts.