Q. 2 C

# Find the equation of the circle passing through the points :

(5, - 8), (- 2, 9) and (2, 1)

Answer :

Given that we need to find the equation of the circle passing through the points (5, - 8), (- 2,9) and (2,1).

We know that the standard form of the equation of a circle is given by:

⇒ x^{2} + y^{2} + 2ax + 2by + c = 0 ..... (1)

Substituting (5, - 8) in (1), we get

⇒ 5^{2} + (- 8)^{2} + 2a(5) + 2b(- 8) + c = 0

⇒ 25 + 64 + 10a - 16b + c = 0

⇒ 10a - 16b + c + 89 = 0 ..... (2)

Substituting (- 2,9) in (1), we get

⇒ (- 2)^{2} + 9^{2} + 2a(- 2) + 2b(9) + c = 0

⇒ 4 + 81 - 4a + 18b + c = 0

⇒ - 4a + 18b + c + 85 = 0 ..... (3)

Substituting (2,1) in (1), we get

⇒ 2^{2} + 1^{2} + 2a(2) + 2b(1) + c = 0

⇒ 4 + 1 + 4a + 2b + c = 0

⇒ 4a + 2b + c + 5 = 0 ..... (4)

Solving (2), (3), (4) we get

⇒ a = 58, b = 24,c = - 285.

Substituting these values in (1), we get

⇒ x^{2} + y^{2} + 2(58)x + 2(24) - 285 = 0

⇒ x^{2} + y^{2} + 116x + 48y - 285 = 0

∴ The equation of the circle is x^{2} + y^{2} + 116x + 48y - 285 = 0.

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