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.

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.

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