Find the equation of the circle passing through the points :
(1, 2), (3, - 4) and (5, - 6)
Given that we need to find the equation of the circle passing through the points (1,2), (3, - 4) and (5, - 6).
We know that the standard form of the equation of a circle is given by:
⇒ x2 + y2 + 2ax + 2by + c = 0 ..... (1)
Substituting (1,2) in (1), we get
⇒ 12 + 22 + 2a(1) + 2b(2) + c = 0
⇒ 1 + 4 + 2a + 4b + c = 0
⇒ 2a + 4b + c + 5 = 0 ..... (2)
Substituting (3, - 4) in (1), we get
⇒ 32 + (- 4)2 + 2a(3) + 2b(- 4) + c = 0
⇒ 9 + 16 + 6a - 8b + c = 0
⇒ 6a - 8b + c + 25 = 0 ..... (3)
Substituting (5, - 6) in (1), we get
⇒ 52 + (- 6)2 + 2a(5) + 2b(- 6) + c = 0
⇒ 25 + 36 + 10a - 12b + c = 0
⇒ 10a - 12b + c + 61 = 0 .....(4)
Solving (2), (3), (4) we get
⇒ a = - 11, b = - 2,c = 25.
Substituting these values in (1), we get
⇒ x2 + y2 + 2(- 11)x + 2(- 2) + 25 = 0
⇒ x2 + y2 - 22x - 4y + 25 = 0
∴ The equation of the circle is x2 + y2 - 22x - 4y + 25 = 0.