For intersection points of lines
x + y = 0
⇒ y = -x
and 2x + y + 5 = 0
⇒ 2x - x + 5 = 0
⇒ x = -5
\(\therefore \) Intersection point is A(-5, 5)
x - y = 0
⇒ y = x
and 2x + y + 5 = 0
⇒ 2x + x + 5 = 0
⇒ \(x = \frac{-5}3\)
\(\therefore \) Intersection point is B\(\left(\frac{-5}3,\frac{-5}3\right)\).
For intersection points x + y = 0 & x - y = 0
x + x = 0
⇒ x = 0
⇒ y = 0
\(\therefore \) Intersection point is C(0, 0).
\(\therefore\) Let the circumcentre be (x, y).
Then
\((x + 5)^2 + (y - 5)^2 = (x + \frac 53)^2 + (y + \frac53)^2\)
\(= x^2 +y^2\)
⇒ \(x^2 + y^2 + 10(x-y) + 50 = x^2 + y^2+\frac{10}3 (x + y) + \frac{50}9\)
\(= x^2 + y^2\)
⇒ \(10(x - y) + 50 = \frac{10}3 (x + y) + \frac{50}9 = 0\)
⇒ \(x-y = 5\;and\; x + y = \frac{-50}9 \times \frac 3{10} = \frac{-5}3\)
⇒ \(2x = -5 - \frac 53 = \frac{-20}3\)
⇒ \(x = \frac{-10}3\)
\(\therefore y=x + 5 = \frac{-10}3 + 5 = \frac 53\)
\(\therefore \) Circumcentre of formed triangle is \(\left(\frac{-10}3, \frac 53\right)\).
