Question -
Answer -
Given:
The points (3, -2),(1, 0), (-1, -2) and (1, -4)
Let us assume thecircle passes through the points A, B, C.
So by using thestandard form of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0….. (1)
Substitute the pointsA (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….. (2)
Substitute the pointsB (1, 0) in equation (1), we get,
12 + 02 +2a(1) + 2b(0) + c = 0
1 + 2a + c = 0 ……- (3)
Substitute the pointsC (-1, -2) in equation (1), we get,
(- 1)2 +(- 2)2 + 2a(- 1) + 2b(- 2) + c = 0
1 + 4 – 2a – 4b + c =0
5 – 2a – 4b + c = 0
2a + 4b – c – 5 = 0…..(4)
Upon simplifying theequations (2), (3) and (4) we get,
a = – 1, b = 2 and c =1
Substituting thevalues of a, b, c in equation (1), we get
x2 + y2 +2(- 1)x + 2(2)y + 1 = 0
x2 + y2 –2x + 4y + 1 = 0 ….. (5)
Now by substitutingthe point D (1, -4) in equation (5) we get,
12 +(- 4)2 – 2(1) + 4(- 4) + 1
1 + 16 – 2 – 16 + 1
0
∴ The points (3,-2), (1, 0), (-1, -2), (1, -4) are con – cyclic.