Question -
Answer -
Given:
The points (5, 5), (6,4), (- 2, 4) and (7, 1) all lie on a circle.
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)
Substituting A (5, 5)in (1), we get,
52 + 52 +2a(5) + 2b(5) + c = 0
25 + 25 + 10a + 10b +c = 0
10a + 10b + c + 50 =0….. (2)
Substitute the pointsB (6, 4) in equation (1), we get,
62 + 42 +2a(6) + 2b(4) + c = 0
36 + 16 + 12a + 8b + c= 0
12a + 8b + c + 52 =0….. (3)
Substitute the point C(-2, 4) in equation (1), we get,
(-2)2 +42 + 2a(-2) + 2b(4) + c = 0
4 + 16 – 4a + 8b + c =0
20 – 4a + 8b + c = 0
4a – 8b – c – 20 =0….. (4)
Upon simplifyingequations (2), (3) and (4) we get,
a = – 2, b = – 1 and c= – 20
Now by substitutingthe values of a, b, c in equation (1), we get
x2 + y2 +2(- 2)x + 2(- 1)y – 20 = 0
x2 + y2 –4x – 2y – 20 = 0 ….. (5)
Substituting D (7, 1)in equation (5) we get,
72 + 12 –4(7) – 2(1) – 20
49 + 1 – 28 – 2 – 20
0
∴ The points (3,-2), (1, 0), (-1, -2), (1, -4) lie on a circle.
Now let us find thecentre and the radius.
We know that for acircle x2 + y2 + 2ax + 2by + c = 0,
Centre = (-a, -b)
Radius = √(a2 + b2 –c)
Comparing equation (5)with equation (1), we get
Centre = [-(-4)/2,-(-2)/2)]
= (2, 1)
Radius = √(22 + 12 –(-20))
= √(25)
= 5
∴ The centre andradius of the circle is (2, 1) and 5.