NCERT Solutions Class 11 Maths Chapter 10 Conic Sections Exercise 10.1

Exercise 10.1

1. Find the equation of the circle with 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

${{(x-0)}^{2}}+{{(y-2)}^{2}}={{2}^{2}}$

$\begin{align}   & {{x}^{2}}+{{y}^{2}}+4-4y=4 \\  & {{x}^{2}}+{{y}^{2}}-4y=0 \\ \end{align}$

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

${{(x+2)}^{2}}+{{(y-3)}^{2}}={{(4)}^{2}}$

$\begin{align}  & {{x}^{2}}+4x+4+{{y}^{2}}-6y+9=16 \\  & {{x}^{2}}+{{y}^{2}}+4x-6y-3=0 \\ \end{align}$

3. Find the equation of the circle with centre $\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ and radius $\left( \dfrac{1}{12} \right)$

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 $(h,k)=\left( \dfrac{1}{2},\dfrac{1}{4} \right)$ and radius $(r)=\left( \dfrac{1}{12} \right).$

Therefore, the equation of the circle is

${{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y-\dfrac{1}{4} \right)}^{2}}={{\left( \dfrac{1}{12} \right)}^{2}}$

${{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}=\dfrac{1}{144}$

${{x}^{2}}-x+\dfrac{1}{4}+{{y}^{2}}-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0$

$144{{x}^{2}}-144x+36+144{{y}^{2}}-72y+9-1=0$ (Solve by taking LCM)

$144{{x}^{2}}-144x+144{{y}^{2}}-72y+44=0$

$36{{x}^{2}}-36x+36{{y}^{2}}-18y+11=0$

$36{{x}^{2}}+36{{y}^{2}}-36x-18y+11=0$

4. Find the equation of the circle with centre (1,1) and radius $\sqrt{2}$

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

${{(x-1)}^{2}}+{{(y-1)}^{2}}={{(\sqrt{2})}^{2}}$

$\begin{align}  & {{x}^{2}}-2x+1+{{y}^{2}}-2y+1=2 \\  & {{x}^{2}}+{{y}^{2}}-2x-2y=0 \\ \end{align}$

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

${{(x+a)}^{2}}+{{(y+b)}^{2}}={{(\sqrt{{{a}^{2}}-{{b}^{2}}})}^{2}}$

$\begin{align}   & {{x}^{2}}+2ax+{{a}^{2}}+{{y}^{2}}+2by+{{b}^{2}}={{a}^{2}}-{{b}^{2}} \\  & {{x}^{2}}+{{y}^{2}}+2ax+2by+2{{b}^{2}}=0 \\ \end{align}$

6. Find the centre and radius of the circle ${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

Ans: The equation of the given circle is ${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

${{(x+5)}^{2}}+{{(y-3)}^{2}}=36$

$\Rightarrow {{\{x-(-5)\}}^{2}}+{{(y-3)}^{2}}={{(6)}^{2}}$, which is the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=-5,k=3,and\; r=6$.

Thus, the centre of the given circle is $(-5,3),$ while its radius is 6.

7. Find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-4x-8y-45=0$

$\Rightarrow ({{x}^{2}}-4x)+({{y}^{2}}-8y)=45$

$\Rightarrow \{{{x}^{2}}-2(x)(2)+{{(2)}^{2}}\}+\{{{y}^{2}}-2(y)(4)+{{(4)}^{2}}\}-4-16=45$

$\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}=65$

$\Rightarrow {{(x-2)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{65} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=2,k=4,and\; r=\sqrt{65}$

Thus, the centre of the given circle is $(2,4),$while its radius is $\sqrt{65}$.

8. Find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-8x+10y-12=0$

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}-8x+10y-12=0$

$\Rightarrow {{x}^{2}}+{{y}^{2}}-8x+10y-12=0$
$\Rightarrow ({{x}^{2}}-8x)+({{y}^{2}}+10y)=12$

$\Rightarrow \{{{x}^{2}}-2(x)(4)+{{(4)}^{2}}\}+\{{{y}^{2}}+2(y)(5)+{{(5)}^{2}}-16-25=12$

$\Rightarrow {{(x-4)}^{2}}+{{(y+5)}^{2}}=53$

$\Rightarrow {{(x-4)}^{2}}+{{\{y-(-5)\}}^{2}}={{\left( \sqrt{53} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=4,k=-5,and\; r=\sqrt{53}$

Thus, the centre of the given circle is $(4,-5),$ while its radius is $\sqrt{53}$.

9. Find the centre and radius of the circle $2{{x}^{2}}+2{{y}^{2}}-x=0$

Ans: The equation of the given circle is $2{{x}^{2}}+2{{y}^{2}}-x=0$

$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}-x=0$
$\Rightarrow (2{{x}^{2}}-x)+2{{y}^{2}}=0$

$\Rightarrow 2\left[ \left( {{x}^{2}}-\dfrac{x}{2} \right)+{{y}^{2}} \right]=0$

$\Rightarrow \left\{ {{x}^{2}}-2.x\left( \dfrac{1}{4} \right)+{{\left( \dfrac{1}{4} \right)}^{2}} \right\}+{{y}^{2}}-{{\left( \dfrac{1}{4} \right)}^{2}}=0$

$\Rightarrow {{\left( x-\dfrac{1}{4} \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \dfrac{1}{4} \right)}^{2}},$ which is of the form ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h=\dfrac{1}{4},k=0,$ and $r = \dfrac{1}{4}$

Thus, the centre of the given circle is $\left( \dfrac{1}{4},0 \right),$ while its radius is $\dfrac{1}{4}$.

10. 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: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(4,1)$ and $(6,5)$,

${{(4-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(i)

${{(6-h)}^{2}}+{{(5-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $4x+y=16,$

$4h+k=16$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(4-h)}^{2}}+{{(1-k)}^{2}}={{(6-h)}^{2}}+{{(5-k)}^{2}}$

$\Rightarrow 16-8h+{{h}^{2}}+1-2k+{{k}^{2}}=36-12h+{{h}^{2}}+25-10k+{{k}^{2}}$

$\begin{align}   & \Rightarrow 16-8h+1-2k=36-12h+25-10k \\  & \Rightarrow 4h+8k=44 \\ \end{align}$

$\Rightarrow h+2k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=3$ and $k=4$

On substituting the values of h and k in equation (i), we obtain

${{(4-3)}^{2}}+{{(1-4)}^{2}}={{r}^{2}}$

$\Rightarrow {{(1)}^{2}}+{{(-3)}^{2}}={{r}^{2}}$

$\Rightarrow 1+9={{r}^{2}}$

$\begin{align} & \Rightarrow {{r}^{2}}=10 \\  & \Rightarrow r=\sqrt{10} \\ \end{align}$

Thus, the equation of the required circle is

${{(x-3)}^{2}}+{{(y-4)}^{2}}={{\left( \sqrt{10} \right)}^{2}}$

$\begin{align}   & {{x}^{2}}-6x+9+{{y}^{2}}-8y+16=10 \\  & {{x}^{2}}+{{y}^{2}}-6x-8y+15=0 \\ \end{align}$

11. 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: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through points $(2,3)$ and $(-1,1)$,

${{(2-h)}^{2}}+{{(3-k)}^{2}}={{r}^{2}}$ ………(i)

${{(-1-h)}^{2}}+{{(1-k)}^{2}}={{r}^{2}}$ ………(ii)

Since the centre (h,k) of the circle lies on line $x-3y-11=0$

$h-3k=11$ ………(iii)

From equations (i) and (ii), we get

$\Rightarrow {{(2-h)}^{2}}+{{(3-k)}^{2}}={{(-1-h)}^{2}}+{{(1-k)}^{2}}$

$\Rightarrow 4-4h+{{h}^{2}}+9-6k+{{k}^{2}}=1+2h+{{h}^{2}}+1-2k+{{k}^{2}}$

$\Rightarrow 4-4h+9-6k=1+2h+1-2k$

$\Rightarrow 6h+4k=11$ ………(iv)

On solving equations (iii) and (iv), we obtain $h=\dfrac{7}{2}$ and $k=\dfrac{-5}{2}$

On substituting the values of h and k in equation (i), we obtain

$\Rightarrow {{\left( 2-\dfrac{7}{2} \right)}^{2}}+{{\left( 3+\dfrac{5}{2} \right)}^{2}}={{r}^{2}}$

$\Rightarrow {{\left( \dfrac{4-7}{2} \right)}^{2}}+{{\left( \dfrac{6+5}{2} \right)}^{2}}={{r}^{2}}$

$\Rightarrow {{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}={{r}^{2}}$

$\begin{align}   & \Rightarrow \dfrac{9}{4}+\dfrac{121}{4}={{r}^{2}} \\  & \Rightarrow \dfrac{130}{4}={{r}^{2}} \\ \end{align}$

Thus, the equation of the required circle is

${{\left( x-\dfrac{7}{2} \right)}^{2}}+{{\left( y+\dfrac{5}{2} \right)}^{2}}=\dfrac{130}{4}$

$\begin{align}   & {{\left( \dfrac{2x-7}{2} \right)}^{2}}+{{\left( \dfrac{2y+5}{2} \right)}^{2}}=\dfrac{130}{4} \\ & 4{{x}^{2}}-28x+49+4{{y}^{2}}+20y+25=130 \\ \end{align}$

$\begin{align}   & 4{{x}^{2}}+4{{y}^{2}}-28x+20y-56=0 \\  & 4({{x}^{2}}+y{}^{2}-7x+5y-14)=0 \\ \end{align}$

${{x}^{2}}+{{y}^{2}}-7x+5y-14=0$

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the radius of the circle  is 5 and its centre lies on the x-axis, k=0 and r=5.

Now, the equation of the circle passes through point $(2,3).$

$\begin{align}   & \therefore {{\left( 2-h \right)}^{2}}+{{3}^{2}}=25 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=25-9 \\  & \Rightarrow {{\left( 2-h \right)}^{2}}=16 \\ \end{align}$

$\begin{align}  & \Rightarrow \left( 2-h \right)=\pm \sqrt{16} \\  & =\pm 4 \\ \end{align}$

If $2-h=4,$ then $h=-2$

If $2-h=-4,$ then $h=6$

When $h=-2$, the equation of the circle becomes

$\begin{align}  & {{\left( x+2 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}+4x+4+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}+4x-21=0 \\ \end{align}$

When $h=6$, the equation of the circle becomes

$\begin{align}   & {{\left( x-6 \right)}^{2}}+{{y}^{2}}=25 \\  & {{x}^{2}}-12x+36+{{y}^{2}}=25 \\  & {{x}^{2}}+{{y}^{2}}-12x+11=0 \\ \end{align}$

So, the equation of the circle can be ${{x}^{2}}+{{y}^{2}}+4x-21=0 $ or ${{x}^{2}}+{{y}^{2}}-12x+11=0 $

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

Ans: Let the equation of the required circle be ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

Since the circle passes through (0,0).

$\begin{align}  & {{(0-h)}^{2}}+{{(0-k)}^{2}}={{r}^{2}} \\  & \Rightarrow {{h}^{2}}+{{k}^{2}}={{r}^{2}} \\ \end{align}$

The equation of the circle now becomes ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{h}^{2}}+{{k}^{2}}$

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points $(a,0)$ and $(0,b)$. Therefore,

${{(a-h)}^{2}}+{{(0-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(1)

${{(0-h)}^{2}}+{{(b-k)}^{2}}={{h}^{2}}+{{k}^{2}}$ …….(2)

From equation (1), we’ll get

$\begin{align}  & {{a}^{2}}-2ah+{{h}^{2}}+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{a}^{2}}-2ah=0 \\  & \Rightarrow a(a-2h)=0 \\ \end{align}$

$\Rightarrow a=0$ or $(a-2h)=0$

However, $a\ne 0;$ hence, $(a-2h)=0\Rightarrow h=\dfrac{a}{2}$

From equation (2), we’ll get

$\begin{align}  & {{h}^{2}}+{{b}^{2}}-2bk+{{k}^{2}}={{h}^{2}}+{{k}^{2}} \\  & \Rightarrow {{b}^{2}}-2bk=0 \\  & \Rightarrow b(b-2k)=0 \\ \end{align}$

$\Rightarrow b=0$ or $(b-2k)=0$

However, $b\ne 0;$ hence, $(b-2k)=0\Rightarrow k=\dfrac{b}{2}$

Thus, the equation of the required circle is

${{\left( x-\dfrac{a}{2} \right)}^{2}}+{{\left( y-\dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{a}{2} \right)}^{2}}+{{\left( \dfrac{b}{2} \right)}^{2}}$

${{\left( \dfrac{2x-a}{2} \right)}^{2}}+{{\left( \dfrac{2y-b}{2} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{4}$

$\begin{align}  & \Rightarrow 4{{x}^{2}}-4ax+{{a}^{2}}+4{{y}^{2}}-4by+{{b}^{2}}={{a}^{2}}+{{b}^{2}} \\  & \Rightarrow 4{{x}^{2}}+4{{y}^{2}}-4ax-4by=0 \\  & \Rightarrow {{x}^{2}}+{{y}^{2}}-ax-by=0 \\ \end{align}$

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

Ans: The centre of the circle is given as (h,k)=(2,2)

Since the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

$\begin{align}  & \therefore r=\sqrt{{{(2-4)}^{2}}+{{(2-5)}^{2}}} \\  & =\sqrt{{{(-2)}^{2}}+{{(-3)}^{2}}} \\  & =\sqrt{4+9} \\  & =\sqrt{13} \\ \end{align}$

Thus, the equation of the circle is 

${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$

$\begin{align}  & {{(x-2)}^{2}}+{{(y-2)}^{2}}={{\left( \sqrt{13} \right)}^{2}} \\  & {{x}^{2}}-4x+4+{{y}^{2}}-4y+4=13 \\  & {{x}^{2}}+{{y}^{2}}-4x-4y-5=0 \\ \end{align}$

15. Does the point (-2.5, 3.5) lie inside, outside or on the circle ${{x}^{2}}+{{y}^{2}}=25$?

Ans: The equation of the given circle is ${{x}^{2}}+{{y}^{2}}=25$.

${{x}^{2}}+{{y}^{2}}=25$

$\Rightarrow {{(x-0)}^{2}}+{{(y-0)}^{2}}={{5}^{2}},$ which is of the form of ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, where $h = 0, k = 0 \; and\; r=5$

$\therefore $ Centre =(0,0) and radius =5

Distance between point (-2.5, 3.5) and centre (0,0)

$=\sqrt{{{(-2.5-0)}^{2}}+{{(3.5-0)}^{2}}}$

$\begin{align}  & =\sqrt{6.25+12.25} \\  & =\sqrt{18.25} \\ \end{align}$

$=4.3$ (approx.) $<5$

Since the distance between point (-2.5, 3.5) and centre (0,0) of the circle is less than the radius of the circle, point (-2.5, 3.5) lies inside the circle.

Class 11 Maths Chapter 10: Exercises Breakdown

Conclusion

NCERT Solutions for Class 11 Maths Chapter 10 - Conic Sections: Exercise 10.1 provide a comprehensive and clear explanation of the fundamental concepts of conic sections. By utilising these step-by-step solutions, students can easily understand the properties and equations of circles, parabolas, ellipses, and hyperbolas. This resource not only helps in solving textbook problems but also strengthens the foundation for future mathematical concepts. Downloading the FREE PDF ensures that students have a reliable tool for homework, revisions, and exam preparations.

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