Find the eq

Given that we need to find the equation of the circle passing through the points (5,7), (8,1) and (1,3).

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

⇒ x² + y² + 2ax + 2by + c = 0 ..... (1)

Substituting (5,7) in (1), we get

⇒ 5² + 7² + 2a(5) + 2b(7) + c = 0

⇒ 25 + 49 + 10a + 14b + c = 0

⇒ 10a + 14b + c + 74 = 0 ..... (2)

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

⇒ 8² + 1² + 2a(8) + 2b(1) + c = 0

⇒ 64 + 1 + 16a + 2b + c = 0

⇒ 16a + 2b + c + 65 = 0 ..... (3)

Substituting (1,3) in (1), we get

⇒ 1² + 3² + 2a(1) + 2b(3) + c = 0

⇒ 1 + 9 + 2a + 6b + c = 0

⇒ 2a + 6b + c + 10 = 0 ..... (4)

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

⇒ .

Substituting these values in (1), we get

⇒

⇒

⇒ 3x² + 3y² - 29x - 19y + 56 = 0

∴ The equation of the circle is 3x² + 3y² - 29x - 19y + 56 = 0.