Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.
Given that we need to find the equation of the circle which passes through (2,3), (4,5) and has its centre on the line y - 4x + 3 = 0. ..... (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,
⇒ - 4(- a) + (- b) + 3 = 0
⇒ 4a - b + 3 = 0 ......(3)
Substituting (2,3) in (2), we get
⇒ 22 + 32 + 2a(2) + 2b(3) + c = 0
⇒ 4 + 9 + 4a + 6b + c = 0
⇒ 4a + 6b + c + 13 = 0 ..... (4)
Substituting (4,5) in (2), we get
⇒ 42 + 52 + 2a(4) + 2b(5) + c = 0
⇒ 16 + 25 + 8a + 10b + c = 0
⇒ 8a + 10b + c + 41 = 0 ..... (5)
Solving (3), (4) and (5) we get,
⇒ a = - 2,b = - 5,c = 25
Substituting these values in (2), we get
⇒ x2 + y2 + 2(- 2)x + 2(- 5)y + 25 = 0
⇒ x2 + y2 - 4x - 10y + 25 = 0
∴ The equation of the circle is x2 + y2 - 4x - 10y + 25 = 0.