Question -
Answer -
(i) (5, 7), (8, 1) and (1, 3)
By using the standardform of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0 ….. (1)
Firstly let us findthe values of a, b and c
Substitute the givenpoint (5, 7) in equation (1), we get
52 + 72 +2a (5) + 2b (7) + c = 0
25 + 49 + 10a + 14b +c = 0
10a + 14b + c + 74 =0….. (2)
By substituting thegiven point (8, 1) in equation (1), we get
82 + 12 +2a (8) + 2b (1) + c = 0
64 + 1 + 16a + 2b + c= 0
16a + 2b + c + 65 =0….. (3)
Substituting the point(1, 3) in equation (1), we get
12 + 32 +2a (1) + 2b (3) + c = 0
1 + 9 + 2a + 6b + c =0
2a + 6b + c + 10 =0….. (4)
Now by simplifying theequations (2), (3), (4) we get the values
a = -29/6, b = -19/6,c = 56/3
Substituting thevalues of a, b, c in equation (1), we get
x2 + y2 +2 (-29/6)x + 2 (-19/6) + 56/3 = 0
x2 + y2 –29x/3 – 19y/3 + 56/3 = 0
3x2 +3y2 – 29x – 19y + 56 = 0
∴ The equation ofthe circle is 3x2 + 3y2 – 29x – 19y + 56 = 0
(ii) (1, 2), (3, – 4) and(5, – 6)
By using the standardform of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0 ….. (1)
Substitute the points(1, 2) in equation (1), we get
12 + 22 +2a (1) + 2b (2) + c = 0
1 + 4 + 2a + 4b + c =0
2a + 4b + c + 5 = 0…..(2)
Substitute the points(3, -4) in equation (1), we get
32 +(- 4)2 + 2a (3) + 2b (- 4) + c = 0
9 + 16 + 6a – 8b + c =0
6a – 8b + c + 25 =0….. (3)
Substitute the points(5, -6) in equation (1), we get
52 +(- 6)2 + 2a (5) + 2b (- 6) + c = 0
25 + 36 + 10a – 12b +c = 0
10a – 12b + c + 61 =0….. (4)
Now by simplifying theequations (2), (3), (4) we get
a = – 11, b = – 2, c =25
Substitute the valuesof a, b and c in equation (1), we get
x2 + y2 +2(- 11)x + 2(- 2) + 25 = 0
x2 + y2 –22x – 4y + 25 = 0
∴ The equation ofthe circle is x2 + y2 – 22x – 4y + 25 = 0
(iii) (5, -8), (-2, 9) and (2, 1)
By using the standardform of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0 ….. (1)
Substitute the point(5, -8) in equation (1), we get
52 +(- 8)2 + 2a(5) + 2b(- 8) + c = 0
25 + 64 + 10a – 16b +c = 0
10a – 16b + c + 89 =0….. (2)
Substitute the points(-2, 9) in equation (1), we get
(- 2)2 +92 + 2a(- 2) + 2b(9) + c = 0
4 + 81 – 4a + 18b + c= 0
-4a + 18b + c + 85 =0….. (3)
Substitute the points(2, 1) in equation (1), we get
22 + 12 +2a(2) + 2b(1) + c = 0
4 + 1 + 4a + 2b + c =0
4a + 2b + c + 5 = 0…..(4)
By simplifyingequations (2), (3), (4) we get
a = 58, b = 24, c = –285.
Now, by substitutingthe values of a, b, c in equation (1), we get
x2 + y2 +2(58)x + 2(24) – 285 = 0
x2 + y2 +116x + 48y – 285 = 0
∴ The equation ofthe circle is x2 + y2 + 116x + 48y – 285 = 0
(iv) (0, 0), (-2, 1) and (-3, 2)
By using the standardform of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0 ….. (1)
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….. (2)
Substitute the points(-2, 1) in equation (1), we get
(- 2)2 +12 + 2a(- 2) + 2b(1) + c = 0
4 + 1 – 4a + 2b + c =0
-4a + 2b + c + 5 =0….. (3)
Substitute the points(-3, 2) in equation (1), we get
(- 3)2 +22 + 2a(- 3) + 2b(2) + c = 0
9 + 4 – 6a + 4b + c =0
-6a + 4b + c + 13 =0….. (4)
By simplifying theequations (2), (3), (4) we get
a = -3/2, b = -11/2, c= 0
Now, by substitutingthe values of a, b, c in equation (1), we get
x2 + y2 +2(-3/2)x + 2(-11/2)y + 0 = 0
x2 + y2 –3x – 11y = 0
∴ The equation ofthe circle is x2 + y2 – 3x – 11y = 0