Question

Find the equation of the circle which circumscribes the triangle formed by the lines 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0

Answer 1

In \[∆\]ABC:
Let AB represent the line 2x + y − 3 = 0.    ...(1)
Let BC represent the line x + y − 1 = 0.      ...(2)
Let CA represent the line 3x + 2y − 5 = 0.  ...(3)

Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, −1).
Intersection point of (2) and (3) is (3, −2).

The coordinates of A, B and C are (1, 1), (2, −1) and (3, −2), respectively.

Let the equation of the circumcircle be 

\[x^2 + y^2 + 2gx + 2fy + c = 0\]

It passes through A, B and C.

∴ \[2 + 2g + 2f + c = 0\]

\[5 + 4g - 2f + c = 0\]

\[13 + 6g - 4f + c = 0\]

\[\therefore g = \frac{- 13}{2}, f = \frac{- 5}{2}, c = 16\]

Hence, the required equation of the circumcircle is 

\[x^2 + y^2 - 13x - 5y + 16 = 0\]