Text Book Back Questions and Answers
Question 1.
Find the equation of the following circles having
(i) the centre (3, 5) and radius 5 units.
(ii) the centre (0, 0) and radius 2 units.
Solution:
(i) Equation of the circle is (x – h)2 + (y – k)2 = r2
Centre (h, k) = (3, 5) and radius r = 5
∴ Equation of the circle is (x – 3)2 + (y – 5)2 = 52
⇒ x2 – 6x + 9 + y2 – 10y + 25 = 25
⇒ x2 + y2 – 6x – 10y + 9 = 0
(ii) Equation of the circle when centre origin (0, 0) and radius r is x2 + y2 = r2
⇒ x2 + y2 = 22
⇒ x2 + y2 = 4
⇒ x2 + y2 – 4 = 0
Question 2.
Find the centre and radius of the circle
(i) x2 + y2 = 16
(ii) x2 + y2 – 22x – 4y + 25 = 0
(iii) 5x2 + 5y2+ 4x – 8y – 16 = 0
(iv) (x + 2) (x – 5) + (y – 2) (y – 1) = 0
Solution:
(i) x2 + y2 = 16
⇒ x2 + y2 = 42
This is a circle whose centre is origin (0, 0), radius 4.
(ii) Comparing x2 + y2 – 22x – 4y + 25 = 0 with general equation of circle x2 + y2 + 2gx + 2fy + c = 0
We get 2g = -22, 2f = -4, c = 25
g = -11, f = -2, c = 25
Centre = (-g, -f) = (11, 2)
(iii) 5x2 + 5y2 + 4x – 8y – 16 = 0
To make coefficient of x2 unity, divide the equation by 5 we get,
Comparing the above equation with x2 + y2 + 2gx + 2fy + c = 0 we get,
(iv) Equation of the circle is (x + 2) (x – 5) + (y – 2) (y – 1) = 0
x2 – 3x – 10 + y2 – 3y + 2 = 0
x2 + y2 – 3x – 3y – 8 = 0
Comparing this with x2 + y2 + 2gx + 2fy + c = 0
We get 2g = -3, 2f = -3, c = -8
Question 3.
Find the equation of the circle whose centre is (-3, -2) and having circumference 16π.
Solution:
Circumference, 2πr = 16π
⇒ 2r = 16
⇒ r = 8
Equation of the circle when centre and radius are known is (x – h)2 + (y – k)2 = r2
⇒ (x + 3)2 + (y + 2)2 = 82
⇒ x2 + 6x + 9 + y2 + 4y + 4 = 64
⇒ x2 + y2 + 6x + 4y + 13 = 64
⇒ x2 + y2 + 6x + 4y – 51 = 0
Question 4.
Find the equation of the circle whose centre is (2, 3) and which passes through (1, 4).
Solution:
Centre (h, k) = (2, 3)
Equation of the circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2
⇒ (x – 2)2 + (y – 3)2 = (√2)2
⇒ x2 – 4x + 4 + y2 – 6y + 9 = 2
⇒ x2 + y2 – 4x – 6y + 11 = 0
Question 5.
Find the equation of the circle passing through the points (0, 1), (4, 3) and (1, -1).
Solution:
Let the required of the circle be x2 + y2 + 2gx + 2fy + c = 0 ……… (1)
It passes through (0, 1)
0 + 1 + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 …….. (2)
Again the circle (1) passes through (4, 3)
42 + 32 + 2g(4) + 2f(3) + c = 0
16 + 9 + 8g + 6f + c = 0
8g + 6f + c = -25 …….. (3)
Again the circle (1) passes through (1, -1)
12 + (-1)2 + 2g(1) + 2f(-1) + c = 0
1 + 1 + 2g – 2f + c = 0
2g – 2f + c = -2 ……… (4)
8g + 6f + c = -25
(4) × 4 subtracting we get, 8g – 8f + 4c = -8
14f – 3c = -17 ………. (5)
14f – 3c = -17
(2) × 3 ⇒ 6f + 3c = -3
Adding we get 20f = -20
f = -1
Using f = -1 in (2) we get, 2(-1) + c = -1
c = -1 + 2
c = 1
Using f = -1, c = 1 in (3) we get
8g + 6(-1)+1 = -25
8g – 6 + 1 = -25
8g – 5 = -25
8g = -20
Question 6.
Find the equation of the circle on the line joining the points (1, 0), (0, 1), and having its centre on the line x + y = 1.
Solution:
Let the equation of the circle be
x2 + y2 + 2gx + 2fy + c = 0 ……… (1)
The circle passes through (1, 0)
12 + 02 + 2g(1) + 2f(0) + c = 0
1 + 2g + c = 0
2g + c = 1 …….. (2)
Again the circle (1) passes through (0, 1)
02 + 12 + 2g(0) + 2f(1) + c = 0
1 + 2f + c = 0
2f + c = -1 ……. (3)
(2) – (3) gives 2g – 2f = 0 (or) g – f = 0 ………. (4)
Given that the centre of the circle (-g, -f) lies on the line x + y = 1
-g – f = 1 …….. (5)
Question 7.
If the lines x + y = 6 and x + 2y = 4 are diameters of the circle, and the circle passes through the point (2, 6) then find its equation.
Solution:
To get coordinates of centre we should solve the equations of the diameters x + y = 6, x + 2y = 4.
x + y = 6 ……. (1)
x + 2y = 4 ………. (2)
(1) – (2) ⇒ -y = 2
y = -2
Using y = -2 in (1) we get x – 2 = 6
x = 8
Centre is (8, -2) the circle passes through the point (2, 6).
Equation of the circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2
⇒ (x – 8)2 + (y + 2)2 = 102
⇒ x2 + y2 – 16x + 4y + 64 + 4 = 100
⇒ x2 + y2 – 16x + 4y – 32 = 0
Question 8.
Find the equation of the circle having (4, 7) and (-2, 5) as the extremities of a diameter.
Solution:
The equation of the circle when entremities (x1, y1) and (x2, y2) are given is (x – x1) (x – x2) + (y – y1) (y – y2) = 0
⇒ (x – 4) (x + 2) + (y – 7) (y – 5) = 0
⇒ x2 – 2x – 8 + y2 – 12y + 35 = 0
⇒ x2 + y2 – 2x – 12y + 27 = 0
Question 9.
Find the Cartesian equation of the circle whose parametric equations are x = 3 cos θ, y = 3 sin θ, 0 ≤ θ ≤ 2π.
Solution:
Given x = 3 cos θ, y = 3 sin θ
Now x2 + y2 = 9 cos2θ + 9 sin2θ
x2 + y2 = 9 (cos2θ + sin2θ)
x2 + y2 = 9 which is the Cartesian equation of the required circle.