Q. 15

# 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.

Answer :

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:

⇒ x^{2} + y^{2} + 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

⇒ 2^{2} + 3^{2} + 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

⇒ 4^{2} + 5^{2} + 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

⇒ x^{2} + y^{2} + 2(- 2)x + 2(- 5)y + 25 = 0

⇒ x^{2} + y^{2} - 4x - 10y + 25 = 0

∴ The equation of the circle is x^{2} + y^{2} - 4x - 10y + 25 = 0.

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