Question -
Answer -
Given:
The points (3, 7), (5,5)
The line x – 4y = 1….(1)
By using the standardform of the equation of the circle:
x2 + y2 +2ax + 2by + c = 0 ….. (2)
Let us substitute thecentre (-a, -b) in equation (1) we get,
(- a) – 4(- b) = 1
-a + 4b = 1
a – 4b + 1 = 0…… (3)
Substitute the points(3, 7) in equation (2), we get
32 + 72 +2a(3) + 2b(7) + c = 0
9 + 49 + 6a + 14b + c= 0
6a + 14b + c + 58 =0….. (4)
Substitute the points(5, 5) in equation (2), we get
52 + 52 +2a(5) + 2b(5) + c = 0
25 + 25 + 10a + 10b +c = 0
10a + 10b + c + 50 =0….. (5)
By simplifyingequations (3), (4) and (5) we get,
a = 3, b = 1, c = – 90
Now, by substitutingthe values of a, b, c in equation (2), we get
x2 + y2 +2(3)x + 2(1)y – 90 = 0
x2 + y2 +6x + 2y – 90 = 0
∴ The equation ofthe circle is x2 + y2 + 6x + 2y – 90 = 0.