Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.
Given that we need to find the equation of the circle which passes through (3,7), (5,5) and has its centre on the line x - 4y = 1. ..... (1)
We know that the standard form of the equation of the circle is given by:
⇒ x2 + y2 + 2ax + 2by + c = 0 .....(2)
Substituting centre (- a, - b) in (1) we get,
⇒ (- a) - 4(- b) = 1
⇒ - a + 4b = 1
⇒ a - 4b + 1 = 0 ......(3)
Substituting (3,7) in (2), we get
⇒ 32 + 72 + 2a(3) + 2b(7) + c = 0
⇒ 9 + 49 + 6a + 14b + c = 0
⇒ 6a + 14b + c + 58 = 0 ..... (4)
Substituting (5,5) in (2), we get
⇒ 52 + 52 + 2a(5) + 2b(5) + c = 0
⇒ 25 + 25 + 10a + 10b + c = 0
⇒ 10a + 10b + c + 50 = 0 ..... (5)
Solving (3), (4) and (5) we get,
⇒ a = 3,b = 1,c = - 90
Substituting these values in (2), we get
⇒ x2 + y2 + 2(3)x + 2(1)y - 90 = 0
⇒ x2 + y2 + 6x + 2y - 90 = 0
∴ The equation of the circle is x2 + y2 + 6x + 2y - 90 = 0.