NCERT Solutions for Class 11 Maths Chapter 11 – Conic Sections Ex 11.1
Page No 241:
Question 1:
Find the equation of the circle with centre (0, 2) and radius 2
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
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 = 22
x2 + y2 + 4 – 4 y = 4
x2 + y2 – 4y = 0
Question 2:
Find the equation of the circle with centre (–2, 3) and radius 4
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
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
x2 + 4x + 4 + y2 – 6y + 9 = 16
x2 + y2 + 4x – 6y – 3 = 0
Question 3:
Find the equation of the circle with centre
and radius![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3dee11fb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_5cf8a60.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3dee11fb.gif)
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) =
and radius (r) =
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_5cf8a60.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3dee11fb.gif)
Therefore, the equation of the circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4745/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_41b950c5.gif)
Question 4:
Find the equation of the circle with centre (1, 1) and radius![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4746/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m79f714c4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4746/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m79f714c4.gif)
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (1, 1) and radius (r) =
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4746/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m79f714c4.gif)
Therefore, the equation of the circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4746/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m4a3eabf1.gif)
Question 5:
Find the equation of the circle with centre (–a, –b) and radius![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4747/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m24fae901.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4747/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m24fae901.gif)
Answer:
The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–a, –b) and radius (r) =
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4747/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m24fae901.gif)
Therefore, the equation of the circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4747/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m464ae8a7.gif)
Question 6:
Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36
Answer:
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 = 62, which is of the form (x – h)2 + (y – k)2 = r2, 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 x2 + y2 – 4x – 8y – 45 = 0
Answer:
The equation of the given circle is x2 + y2 – 4x – 8y – 45 = 0.
x2 + y2 – 4x – 8y – 45 = 0
⇒ (x2 – 4x) + (y2 – 8y) = 45
⇒ {x2 – 2(x)(2) + 22} + {y2 – 2(y)(4)+ 42} – 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 = r2, where h = 2, k = 4, and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4749/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_75976ae5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4749/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_29327b20.gif)
Thus, the centre of the given circle is (2, 4), while its radius is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4749/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4d84506b.gif)
Question 8:
Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0
Answer:
The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.
x2 + y2 – 8x + 10y – 12 = 0
⇒ (x2 – 8x) + (y2 + 10y) = 12
⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12
⇒ (x – 4)2 + (y + 5)2 = 53
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5438/chapter%2011_html_6572a691.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5438/chapter%2011_html_m4cf8bd3c.gif)
Thus, the centre of the given circle is (4, –5), while its radius is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5438/chapter%2011_html_5a6990f0.gif)
Question 9:
Find the centre and radius of the circle 2x2 + 2y2 – x = 0
Answer:
The equation of the given circle is 2x2 + 2y2 – x = 0.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m349d3c01.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4a9ba61a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_682bd651.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3783feec.gif)
Thus, the centre of the given circle is
, while its radius is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m18e9ac6f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4751/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m3c88992d.gif)
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.
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)2 + (1 – k)2 = r2 … (1)
(6 – h)2 + (5 – k)2 = r2 … (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 + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2
⇒ 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 = r2
⇒ (1)2 + (– 3)2 = r2
⇒ 1 + 9 = r2
⇒ r2 = 10
⇒ ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4774/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m77cb47e4.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4774/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m77cb47e4.gif)
Thus, the equation of the required circle is
(x – 3)2 + (y – 4)2 = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4774/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m74589c29.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4774/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m74589c29.gif)
x2 – 6x + 9 + y2 – 8y + 16 = 10
x2 + y2 – 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.
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the circle passes through points (2, 3) and (–1, 1),
(2 – h)2 + (3 – k)2 = r2 … (1)
(–1 – h)2 + (1 – k)2 = r2 … (2)
Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,
h – 3k = 11 … (3)
From equations (1) and (2), we obtain
(2 – h)2 + (3 – k)2 = (–1 – h)2 + (1 – k)2
⇒ 4 – 4h + h2 + 9 – 6k + k2 = 1 + 2h + h2 + 1 – 2k + k2
⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k
⇒ 6h + 4k = 11 … (4)
On solving equations (3) and (4), we obtain
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5439/chapter%2011_html_4ea31d7f.gif)
On substituting the values of h and k in equation (1), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5439/chapter%2011_html_2c9eb823.gif)
Thus, the equation of the required circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/5439/chapter%2011_html_m6af5ad29.gif)
Question 12:
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Answer:
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
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 + y2 = 25.
It is given that the circle passes through point (2, 3).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4777/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m580764ba.gif)
When h = –2, the equation of the circle becomes
(x + 2)2 + y2 = 25
x2 + 4x + 4 + y2 = 25
x2 + y2 + 4x – 21 = 0
When h = 6, the equation of the circle becomes
(x – 6)2 + y2 = 25
x2 – 12x +36 + y2 = 25
x2 + y2 – 12x + 11 = 0
Question 13:
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Answer:
Let the equation of the required circle be (x - h)2 + (y - k)2 = r2.
Since the circle passes through (0, 0),
(0 - h)2 + (0 - k)2 = r2
-> h2 + k2 = r2
The equation of the circle now becomes (x - h)2 + (y - k)2 = h2 + k2.
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 = h2 + k2 -> (1)
(0 - h)2 + (b - k)2 = h2 + k2 -> (2)
From equation (1), we obtain
a2 - 2ah + h2 + k2 = h2 + k2
-> a2 - 2ah = 0
-> a(a - 2h) = 0
-> a = 0 or (a - 2h) = 0
However, a not equal to 0; hence, (a - 2h) = 0 -> h =
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4778/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4a974ada.gif)
From equation (2), we obtain
h2 + b2 - 2bk + k2 = h2 + k2
-> b2 - 2bk = 0
-> b(b - 2k) = 0
-> b = 0 or(b - 2k) = 0
However, b not equal to 0; hence, (b - 2k) = 0 -> k =
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4778/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_m7a5d0515.gif)
Thus, the equation of the required circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4778/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_32df6a61.gif)
Question 14:
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Answer:
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).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4779/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_3baf47e4.gif)
Thus, the equation of the circle is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4779/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_4e87d43a.gif)
Question 15:
Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Answer:
The equation of the given circle is x2 + y2 = 25.
x2 + y2 = 25
⇒ (x – 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2 = r2, 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)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/171/4780/NS_15-10-08_KHushboo_11_Math_Chapter11.1_15_SU_SNK_html_61a5f762.gif)
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.