(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3
Given:
The lines x + y + 3 = 0
x – y + 1 = 0
x = 3
On solving these lines we get the intersection points A (-2, -1), B (3, 4), C (3, -6)
So by using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (-2, -1) in equation (1), we get
(- 2)2 + (- 1)2 + 2a(-2) + 2b(-1) + c = 0
4 + 1 – 4a – 2b + c = 0
5 – 4a – 2b + c = 0
4a + 2b – c – 5 = 0….. (2)
Substitute the points (3, 4) in equation (1), we get
32 + 42 + 2a(3) + 2b(4) + c = 0
9 + 16 + 6a + 8b + c = 0
6a + 8b + c + 25 = 0….. (3)
Substitute the points (3, -6) in equation (1), we get
32 + (- 6)2 + 2a(3) + 2b(- 6) + c = 0
9 + 36 + 6a – 12b + c = 0
6a – 12b + c + 45 = 0….. (4)
Upon simplifying equations (2), (3), (4) we get
a = – 3, b = 1, c = -15.
Now by substituting the values of a, b, c in equation (1), we get
x2 + y2 + 2(- 3)x + 2(1)y – 15 = 0
x2 + y2 – 6x + 2y – 15 = 0
∴ The equation of the circle is x2 + y2 – 6x + 2y – 15 = 0.
(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0
Given:
The lines 2x + y – 3 = 0
x + y – 1 = 0
3x + 2y – 5 = 0
On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1)
So by using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (2, -1) in equation (1), we get
22 + (- 1)2 + 2a(2) + 2b(- 1) + c = 0
4 + 1 + 4a – 2b + c = 0
4a – 2b + c + 5 = 0….. (2)
Substitute the points (3, -2) in equation (1), we get
32 + (- 2)2 + 2a(3) + 2b(- 2) + c = 0
9 + 4 + 6a – 4b + c = 0
6a – 4b + c + 13 = 0….. (3)
Substitute the points (1, 1) in equation (1), we get
12 + 12 + 2a(1) + 2b(1) + c = 0
1 + 1 + 2a + 2b + c = 0
2a + 2b + c + 2 = 0….. (4)
Upon simplifying equations (2), (3), (4) we get
a = -13/2, b = -5/2, c = 16
Now by substituting the values of a, b, c in equation (1), we get
x2 + y2 + 2 (-13/2)x + 2 (-5/2)y + 16 = 0
x2 + y2 – 13x – 5y + 16 = 0
∴ The equation of the circle is x2 + y2 – 13x – 5y + 16 = 0
(iii) x + y = 2, 3x – 4y = 6 and x – y = 0
Given:
The lines x + y = 2
3x – 4y = 6
x – y = 0
On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1)
So by using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (2, 0) in equation (1), we get
22 + 02 + 2a(2) + 2b(0) + c = 0
4 + 4a + c = 0
4a + c + 4 = 0….. (2)
Substitute the point (-6, -6) in equation (1), we get
(- 6)2 + (- 6)2 + 2a(- 6) + 2b(- 6) + c = 0
36 + 36 – 12a – 12b + c = 0
12a + 12b – c – 72 = 0….. (3)
Substitute the points (1, 1) in equation (1), we get
12 + 12 + 2a(1) + 2b(1) + c = 0
1 + 1 + 2a + 2b + c = 0
2a + 2b + c + 2 = 0….. (4)
Upon simplifying equations (2), (3), (4) we get
a = 2, b = 3, c = – 12.
Substituting the values of a, b, c in equation (1), we get
x2 + y2 + 2(2)x + 2(3)y – 12 = 0
x2 + y2 + 4x + 6y – 12 = 0
∴ The equation of the circle is x2 + y2 + 4x + 6y – 12 = 0
(iv) y = x + 2, 3y = 4x and 2y = 3x
Given:
The lines y = x + 2
3y = 4x
2y = 3x
On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)
So by using the standard form of the equation of the circle:
x2 + y2 + 2ax + 2by + c = 0….. (1)
Substitute the points (6, 8) in equation (1), we get
62 + 82 + 2a(6) + 2b(8) + c = 0
36 + 64 + 12a + 16b + c = 0
12a + 16b + c + 100 = 0…… (2)
Substitute the points (0, 0) in equation (1), we get
02 + 02 + 2a(0) + 2b(0) + c = 0
0 + 0 + 0a + 0b + c = 0
c = 0….. (3)
Substitute the points (4, 6) in equation (1), we get
42 + 62 + 2a(4) + 2b(6) + c = 0
16 + 36 + 8a + 12b + c = 0
8a + 12b + c + 52 = 0….. (4)
Upon simplifying equations (2), (3), (4) we get
a = – 23, b = 11, c = 0
Now by substituting the values of a, b, c in equation (1), we get
x2 + y2 + 2(- 23)x + 2(11)y + 0 = 0
x2 + y2 – 46x + 22y = 0
∴ The equation of the circle is x2 + y2 – 46x + 22y = 0
