NCERT Solutions For Class 11 Maths chapter-11 Conic Sections Exercise 11.1
NCERT Solutions for Class-11 Maths Chapter-11 Conic Sections
NCERT Solutions for Class 11 Maths Chapter-11 Conic Sections Exercise 11.1 is Prepared by expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download and share NCERT Solutions for Class 11 Maths.
NCERT Solutions for Class-11 Maths Exercise 11.1
In each of the following Exercises 1 to 5, find the equation of the circle with:
Question 1. 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 = 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
x 2 + y 2 + 4 – 4 y = 4
x 2 + y 2 – 4y = 0
Question 2. 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 = 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
x 2 + 4x + 4 + y 2 – 6y + 9 = 16
x 2 + y 2 + 4x – 6y – 3 = 0
Question
3. Centre
and radius
Solution :
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
) =
and radius (
r
) =
.
Therefore, the equation of the circle is
Question 4. Find the equation of the circle with centre (1,1) 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 = r 2
It is given that centre (h, k) = (1, 1) and radius (r) = √
Therefore, the equation of the circle is
(x−1) 2 +(y−1) 2 = (√2)2x 2 − 2x + 1 + y 2 − 2y + 1
= 2x 2 + y 2 − 2x− 2y=0
Question
5. Centre (–
a
, –
b
) and radius
Solution :
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
) =
.
Therefore, the equation of the circle is
In each of the following Exercises 6 to 9, find the centre and radius of the circles.
Question 6. Find the centre and radius of the circle (x + 5) 2 + (y – 3) 2 = 36
Solution :
The equation of the given circle is ( x + 5) 2 + ( y – 3) 2 = 36.
( x + 5) 2 + ( y – 3) 2 = 36
⇒ { x – (–5)} 2 + ( y – 3) 2 = 6 2 , which is of the form ( 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.
Question 7. Find the centre and radius of the circle x 2 + y 2 – 4x – 8y – 45 = 0
Solution :
The equation of the given circle is x
2
+ y
2
– 4x – 8y – 45 = 0.
x 2 + y 2 – 4x – 8y – 45 = 0
⇒ (x 2 – 4x) + (y 2 – 8y) = 45
⇒ {x 2 – 2(x)(2) + 2 2 } + {y 2 – 2(y)(4)+ 4 2 } – 4 –16 = 45
⇒ (x – 2) 2 + (y –4) 2 = 65
⇒ (x – 2) 2 + (y –4) 2
=
, which is of the form (x – h)
2
+ (y – k)
2
= r
2
, where h = 2, k = 4, and
.
Thus, the centre of the given circle is (2, 4), while its radius
.
Question 8. Find the centre and radius of the circle x 2 + y 2 – 8x + 10y – 12 = 0
Solution :
The equation of the given circle is x
2
+ y
2
– 8x + 10y – 12 = 0.
x 2 + y 2 – 8x + 10y – 12 = 0
⇒ (x 2 – 8x) + (y 2 + 10y) = 12
⇒ {x 2 – 2(x)(4) + 4 2 } + {y 2 + 2(y)(5) + 5 2 }– 16 – 25 = 12
⇒ (x – 4) 2 + (y + 5) 2 = 53
, which is of the form (x – h)
2
+ (y – k)
2
= r
2
, where h = 4, k = –5, and
.
Thus, the centre of the given circle is (4, –5), while its radius is
.
Question 9.Find the centre and radius of the circle 2x 2 + 2y 2 – x = 0
Solution :
The equation of the given circle is 2x
2
+ 2y
2
– x = 0.
, which is of the form (
x
–
h
)
2
+ (
y
–
k
)
2
=
r
2
, where
h
= 1/4,
k
= 0, and
r = 1/4.
Thus, the centre of the given circle is
, while its radius is
1/4
.
Question 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.
Solution :
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 … (1)
(6 – h) 2 + (5 – k) 2 = r 2 … (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 + h 2 + 1 – 2k + k 2 = 36 – 12h + h 2 + 25 – 10k + k 2
⇒ 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 = r 2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r 2 = 10
Thus, the equation of the required circle is
(x – 3)
2
+ (y – 4)
2
=
x 2 – 6x + 9 + y 2 – 8y + 16 = 10
x 2 + y 2 – 6x – 8y + 15 = 0
Question 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.
Solution :
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 … (1)
(–1 – h ) 2 + (1 – k ) 2 = r 2 … (2)
Since the centre ( h , k) of the circle lies on line x – 3 y – 11 = 0,
h – 3 k = 11 … (3)
From equations (1) and (2), we obtain
(2 – h ) 2 + (3 – k ) 2 = (–1 – h ) 2 + (1 – k ) 2
⇒ 4 – 4 h + h 2 + 9 – 6 k + k 2 = 1 + 2 h + h 2 + 1 – 2 k + k 2
⇒ 4 – 4 h + 9 – 6 k = 1 + 2 h + 1 – 2 k
⇒ 6 h + 4 k = 11 … (4)
On solving equations (3) and (4), we obtain
.
On substituting the values of h and k in equation (1), we obtain
Thus, the equation of the required circle is
Question 12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution :
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 becomes ( x – h ) 2 + y 2 = 25.
It is given that the circle passes through point (2, 3).
When h = –2, the equation of the circle becomes
( x + 2) 2 + y 2 = 25
x 2 + 4 x + 4 + y 2 = 25
x 2 + y 2 + 4 x – 21 = 0
When h = 6, the equation of the circle becomes
( x – 6) 2 + y 2 = 25
x 2 – 12 x +36 + y 2 = 25
x 2 + y 2 – 12 x + 11 = 0
Question 13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Solution :
Let the equation of the required circle be (x – h)
2
+ (y – k)
2
= r
2
.
Since the circle passes through (0, 0),
(0 – h) 2 + (0 – k) 2 = r 2
⇒ h 2 + k 2 = r 2
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 obtain
a 2 – 2ah + h 2 + k 2 = h 2 + k 2
⇒ a 2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h = a/2.
From equation (2), we obtain
h 2 + b 2 – 2bk + k 2 = h 2 + k 2
⇒ b 2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or (b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k = b/2.
Thus, the equation of the required circle is
Question 14. Find the equation of the circle with centre (2, 2) and passes through the point (4, 5).
Solution :
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).
Thus, the equation of the circle is
Question 15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25?
Solution :
The equation of the given circle is x
2
+ y
2
= 25.
x 2 + y 2 = 25
⇒ (x – 0) 2 + (y – 0) 2 = 5 2 , which is of the form (x – h) 2 + (y – k) 2 = r 2 , where h = 0, k = 0, and r = 5.
∴Centre = (0, 0) and radius = 5
Distance between point (–2.5, 3.5) and centre (0, 0)
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.
