Find the equation of circle which passes through the points (2,-6) (6,4) and (-3,1)
Answers
Answer:
13x² + 13y² - 1500x + 10y - 9540 = 0
Step-by-step explanation:
General equation of a circle:
x² + y² + Dx + Ey + F = 0
At point (2, -6)
(2)² + (-6)² + (2)D + (-6)E + F = 0
4 + 36 + 2D - 6E + F = 0
40 + 2D - 6E + F = 0
At point (6, 4):
(6)² + (4)² + (6)D + (4)E + F = 0
36 + 16 + 6D + 4E + F = 0
52 + 6D + 4E + F = 0
At point (-3, 1):
(-3)² + (1)² + (-3)D + (1)E + F = 0
9 + 1 - 3D + E + F = 0
10 - 3D + E + F = 0
Solve D, E, F:
40 + 2D - 6E + F = 0 ----------- [ 1 ]
52 + 6D + 4E + F = 0 ----------- [ 2 ]
10 - 3D + E + F = 0 . ----------- [ 3 ]
[ 2 ] - [ 1 ] : 12 + 4D + 10E = 0 ----------- [ 4 ]
[ 1 ] - [ 3 ] : 30 + 5D - 7E = 0 ----------- [ 5 ]
[ 4 ] x 5 : 60 + 20D + 50E = 0 ----------- [ 6 ]
[ 5 ] x 4 : 120 + 20D - 28E = 0 ----------- [ 7 ]
[ 6 ] - [ 7 ]:
- 60 + 78E = 0
78E = 60
E = 10/13
Sub E = 10/13 into [ 6]:
60 + 20D + 50(10/13) = 0
20D = - 30000/13
D = -1500/13
Sub E = 10/13 and D = -1500/13 into [ 1 ]:
40 + 2(10/13) - 6(-1500/13) + F = 0
9540/13 + F = 0
F = - 9540/13
Form the equation:
x² + y² + Dx + Ey + F = 0
x² + y² -1500/13 x + 10/13 y - 9540/13 = 0
13x² + 13y² - 1500x + 10y - 9540 = 0
Answer: 13x² + 13y² - 1500x + 10y - 9540 = 0