Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.1 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 11 Maths Chapter 11 Conic Sections Ex 11.1
In each of the following questions 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. Centre (\(\frac{1}{2}\), \(\frac{1}{4}\) and radius = \(\frac{1}{12}\).
4. Centre (1, 1) and radius \(\sqrt{2}\).
5. Centre (- a, – b) and radius \(\sqrt{a^{2}-b^{2}}\).
Solutions to questions 1 to 5:
1. Centre of the circle (0, 2) and radius = 2.
∴ Equation of the circle is
(x – 0)2 + (y – 2)2 = 22.
or x2 + y2 – 4y = 0.
2. Centre of the circle is (- 2, 3) and radius is 4.
∴ Equation of the circle is
(x + 2)2 + (y – 3)2 = 42.
or x2 + y2 + 4x – 6y + 4 + 9 – 16 = 0 or x2 + y2 + 4x – 6y – 3 = 0.
3. Centre of circle is (\(\frac{1}{2}\), \(\frac{1}{4}\) and radius is \(\frac{1}{12}\).
∴ Equation of the circle is
(x – \(\frac{1}{2}\))2 + (y – \(\frac{1}{4}\))2 = \(\frac{1}{12^{2}}\) = \(\frac{1}{144}\).
or 144x2 + 144y2 – 144x – 72y + 45 = 1.
or 144x2 = 144y2 – 144x – 72y + 44 = 0.
or 36x2 + 36y2 – 36x – 18y + 11 = 0.
4. Centre of circle is (1, 1) and radius = \(\sqrt{2}\)
∴ Equation of circle is
(x – 1)2 + (y – 1)2 = (\(\sqrt{2}\))2 = 2
or x2 + y2 – 2x – 2y + 2 = 2
or x2 + y2 – 2x – 2y = 0.
5. Centre of the circle is (- a, – b) and radius = \(\sqrt{a^{2}-b^{2}}\)
∴ Equation of circle is
(x + a)2 + (y + b)2 = a2 – b2.
or x2 + y2 + 2xa + 2yb + a2 + b2 = a2 – b2.
or x2 + y2 + 2ax + 2by + 2b2 = 0.
In each of the following questions 6 to 9, find the centre and radius of the circle:
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.
Solutions to questions 6 to 9:
6. Comparing the equation of the circle
(x + 5)2 + (y – 3)2 = 36
with (x – h)2 + (y – k)2 = r2, we get
– h = 5 or h = – 5, k = 3, r2 = 36 or r = 6.
∴ Centre of the circle is (- 5, 3) and radius = 6.
7. Equation of the circle is
x2 + y2 – 4x – 8y = 45.
or (x2 – 4x) + (y2 – 8y) = 45
or (x2 – 4x + 4) + (y2 – 8y + 16) = 45 + 4 + 16 = 65.
⇒ (x – 2)2 + (y – 4)2 = 65
∴ Centre is (2, 4) and radius = \(\sqrt{65}\).
8. The equation of the circle is
x2 + y2 – 8x + 10y – 12
or (x2 – 8x) + (y2 + 10y) = 12.
or (x2 – 8x + 16) + (y2 + 10y + 25) = 12 + 16 + 25.
or (x – 4)2 + (y + 5)2 = 53.
∴ Centre is (4, – 5) and radius is \(\sqrt{53}\).
9. Equation of the circle is
2x2 + 2y2 – x = 0
or x2 + y2 – \(\frac{x}{2}\) = 0.
∴ Centre is (\(\frac{1}{4}\), 0) and radius = \(\frac{1}{4}\).
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 circle be
(x – h)2 + (y – k)2 = r2 ………………. (1)
The points (4, 1) and (6, 5) lies on it.
∴ (4 – h)2 + (1 – k)2 = r2
or h2 + k2 – 8h – 2k + 17 = r2 …………….. (2)
and (6 – h)2 + (5 – k)2 = r2
or h2 + k2 – 12h – 10k + 61 = r2 ………………….. (3)
The centre (h, k) lies on
4x + y = 16.
∴ 4h + k = 16 ………………….. (4)
Subtracting (3) from (2), we get
4h + 8k – 44
or h + 2k = 11 ……………………. (5)
Multiplying (5) by 4, we get
4h + 8k = 44.
Subtracting equation (4) from it, we get
7k = 44 – 16 = 28. ⇒ k = 4.
From (5), h + 8 = 11. ⇒ h = 3.
Putting h = 3, k = 4 in (2), we get
9 + 16 – 24 – 8 + 17 = r2
or 42 – 32 = r2
∴ r2 = 10.
∴ Equation of the circle is
(x – 3)2 + (y – 4)2 = 10
or 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.
Solution:
Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 ………………. (1)
The points (2, 3) and (- 1, 1) lies on it.
∴ (2 – h)2 + (3 – k)2 = r2
or h2 + k2 – 4h – 6k + 13 = r2 …………………… (2)
and (- 1 – h)2 + (1 – k)2 = r2 …………………. (3)
Centre (h, k) lies on x – 3y – 11 = 0.
∴ h – 3k – 11 = 0 …………….. (4)
Subtracting (2) from (3), we get
6h + 4k – 11 = 0 ………………. (5)
Multiplying (4) by 6, we get
6h – 18k – 66 = 0 …………………. (6)
Subtracting (6) from (5), we get
22k+ 55 = 0.
∴ k = – \(\frac{55}{22}\) = – \(\frac{5}{2}\).
From (4), h = 3k + 11 = – \(\frac{15}{2}\) + 11 = \(\frac{7}{2}\).
Putting the values of h and k in (2 – h)2 + (3 – k)2 = r2, we get
or x2 + y2 – 7x + 5y – 14 = 0.
Question 12.
Find the equation of the circle wi th radius 5, whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Let the equation of the circle be
(x – h)2 + (y – k)2 = r2 ……………………. (1)
Here, r = 5, So, r2 = 25.
Centre lies onx-axis, i.e.,k = 0.
∴ Eq. (1) becomes
(x – h)2 + y2 = 25.
(2, 3) lies on it.
∴ (2 – h)2 + 9 = 25
or (2 – h)2 = 16
∴ 2 – h = ± 4
∴ h = – 2, 6
When h = – 2, equation of circle is
(x + 2)2 + y2 = 25
or x2 + y2 + 4x – 21 = 0.
When h = 6, equation of the circle is
(x – 6)2 + y2 = 25
x2 + y2 – 12x + 11 = 0.
Thus, required circles are
x2 + y2 + 4x – 21 = 0 and 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 co-ordinate axes.
Solution:
a and b are the intercepts made by the circle on the co-ordinate axes at A and B. C, the mid-point of AB is the centre of the circle.
∴ Centre (\(\frac{a}{2}\), \(\frac{b}{2}\)).
Question 14.
Find the equation of a circle with centre (2, 2) and which passes through the point (4, 5).
Solution:
Centre of the circle C(2, 2).
P(4, 5) is a point on the circle.
∴ Radius = CP = \(\sqrt{(4-2)^{2}+(5-2)^{2}}\)
= \(\sqrt{4+9}\) = \(\sqrt{13}\).
∴ Equation of the circle is
(x – 2)2 + (y – 2)2 = 13.
or x2 + y2 – 4x – 4y = 5.
Question 15.
Does the point (- 2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?
Solution:
Centre C of the circle is (0, 0) and radius = 5.
The point is P(- 2.5, 3.5).
∴ CP < Radius.
∴ P lies inside the circle.