Q. 2 B4.0( 4 Votes )

# Find the equation of the circle passing through the points :(1, 2), (3, - 4) and (5, - 6)

Given that we need to find the equation of the circle passing through the points (1,2), (3, - 4) and (5, - 6).

We know that the standard form of the equation of a circle is given by:

x2 + y2 + 2ax + 2by + c = 0 ..... (1)

Substituting (1,2) in (1), we get

12 + 22 + 2a(1) + 2b(2) + c = 0

1 + 4 + 2a + 4b + c = 0

2a + 4b + c + 5 = 0 ..... (2)

Substituting (3, - 4) in (1), we get

32 + (- 4)2 + 2a(3) + 2b(- 4) + c = 0

9 + 16 + 6a - 8b + c = 0

6a - 8b + c + 25 = 0 ..... (3)

Substituting (5, - 6) in (1), we get

52 + (- 6)2 + 2a(5) + 2b(- 6) + c = 0

25 + 36 + 10a - 12b + c = 0

10a - 12b + c + 61 = 0 .....(4)

Solving (2), (3), (4) we get

a = - 11, b = - 2,c = 25.

Substituting these values in (1), we get

x2 + y2 + 2(- 11)x + 2(- 2) + 25 = 0

x2 + y2 - 22x - 4y + 25 = 0

The equation of the circle is x2 + y2 - 22x - 4y + 25 = 0.

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