Math, asked by nidhigupta28935, 1 year ago

Find the equation of circle which passes through the points (2,-6) (6,4) and (-3,1)

Answers

Answered by TooFree
4

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

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