Conic Sections - Solutions

 CBSE Class–11 Mathematics

NCERT Solutions
Chapter - 11 Conic Sections
Exercise 11.1

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In each of the following Exercises 1 to 5, find the equation of the circle with:

1. Centre (0, 2) and radius 2.

Ans. Given:  and 

Equation of the circle;

 

 

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2. Centre  and radius 4.

Ans. Given:  and 

Equation of the circle ;

 

x2+y2+4x−6y−3=0$x^2+y^2+4x-6y-3=0$

 

3. Centre  and radius 

Ans. Given:  and 

Equation of the circle ;

 

 

 

 

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4. Centre  and radius 

Ans. Given:  and 

Equation of the circle;

 

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5. Centre  and radius 

Ans. Given:  and 

Equation of the circle ;

 

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

6. 

Ans. Given: Equation of the circle;

 

  ……….(i)

On comparing eq. (i) with 

 and 

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7. 

Ans. Given: Equation of the circle: 

 

 

=>(x−2)2+(y−4)2=(65−−√)2 .......(i)$=>(x-2{)}^2+(y-4{)}^2=(\sqrt{65}{)}^2.......(i)$

On comparing eq. (i) with 

 and 

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8. 

Ans. Given: Equation of the circle;

 =>(x2−8x)+(y2−10y)=12$=>(x^2-8x)+(y^2-10y)=12$

=>(x2−8x+42)+(y2−10y+52)=12+42+52$=>(x^2-8x+4^2)+(y^2-10y+5^2)=12+4^2+5^2$

=>(x−4)2+(y−5)2=(53−−√)2$=>(x-4{)}^2+(y-5{)}^2=(\sqrt{53}{)}^2$ .......................... (i)

On comparing eq. (i) with  (x−h)2+(y−k)2=r2$(x-h{)}^2+(y-k{)}^2=r^2$

We get, h = 4 , k = 5 and r=53−−√$r=\sqrt{53}$

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9. 

Ans. Given: Equation of the circle: 

        (divide the equation by 2)

 

 

 (x−14)2+(y−0)2=(14)2$(x-\frac{1}{4}{)}^2+(y-0{)}^2=(\frac{1}{4}{)}^2$…….(i)

On comparing eq. (i) with 

 and 

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10. Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre lies on the line 

Ans. The equation of the circle is  ……….(i)

Circle passes through point (4, 1)

 ……….(ii)

Again Circle passes through point (6, 5)

 ……….(iii)

From eq. (ii) and (iii), we have

 

 ……….(iv)

Since the centre  of the circle lies on the line 

 ……….(v)

On solving eq. (iv) and (v), we have 

Putting the values of  and  in eq. (ii), we have

 

Therefore, the equation of the required circle is

 

 

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11. Find the equation of the circle passing through the points (2, 3) and  and whose centre lies on the line 

Ans. The equation of the circle is  ……….(i)

Circle passes through point (2, 3)

 ……….(ii)

Again Circle passes through point (–1, 1)

 ……….(iii)

From eq. (ii) and (iii), we have

=>h2+k2−4h−6k+13=h2+k2+2h−2k+2$=>h^2+k^2-4h-6k+13=h^2+k^2+2h-2k+2$

 ……….(iv)

Since the centre  of the circle lies on the line x−3y−11=0$x-3y-11=0$

……….(v)

On solving eq. (iv) and (v), we have 

Putting the values of  and  in eq. (ii), we have

 

 

Therefore, the equation of the required circle is

 

 

 

 

 

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

Ans. Since the centre of circle lies on axis, therefore the coordinates of centre is 

Now the circle passes through the point (2, 3). According to the question,

 

 

 

  or 

Taking , Equation of the circle is 

 

 

Taking , Equation of the circle is 

 

 

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

Ans. The circle makes intercepts  with axis and  with axis.

OA =  and OB = 

Coordinates of A and B are  and  respectively.

Now the circle passes through the points O (0, 0), A  and B.

Putting these coordinates of three points in the equation of the circle,

 ………(i)

Circle passing through (0,0)

 

 the circle also passes through (a,0) and (0,b)

 

And 

 

 

Putting the values of  and  in eq. (i), we have

 

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

(4, 5).

Ans. The equation of the circle is ……….(i)

Since the circle passes through point (4, 5) and coordinates of centre are (2, 2).
Radius of circle =  =  = 
Therefore, the equation of the required circle is

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15. Does the point  lie inside, outside or on the circle  ?

Ans. Given: Equation of the circle 

On comparing with , we have  and 
Now distance of the point  from the centre (0, 0)
=  =  =  = 4.3 <
Therefore, the point  lies inside the circle.