The circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(3, 0), B(–1, –6), and C(4, –1) be the vertices of ΔABC and P(x, y) be the circumcentre of this triangle. Then, PA = PB = PC ⇒ PA2 = PB2 = PC2 Now, PA2 = PB2 ⇒ (x – 3)2 + (y – 0)2 = (x + 1)2 + (y + 6)2 ⇒ x2 – 6x + 9 + y2 = x2 + 2x + 1 + y2 + 12y + 36 ⇒ – 6x + 9 – 2x – 1 – 12y – 36 = 0 ⇒ 8x + 12y + 28 = 0 ⇒ 2x + 3y = – 7 ...(i) PB2 = PC2 ⇒ (x + 1)2 + (y + 6)2 = (x – 4)2 + (y + 1)2 ⇒ x2 + 2x + 1 + y2 + 12y + 36 = x2 – 8x + 16 + y2 + 2y + 1 ⇒ 10x + 10y + 20 = 0 ⇒ x + y = –2 ...(ii) Solving eqn (i) and eqn (ii) simultaneously, we have (i) – 2 × (ii) ⇒ (2x + 3y) – (2x + 2y) = –7 + 4 ⇒ y = –3 ⇒ x – 3 = –2 ⇒ x = 1 ∴ Co-ordinates of circumcentre P ≡ (1, –3)∴ Circumradius = PA = PB = PC = \(\sqrt{(1-3)^2+(-3-0)^2}\) = \(\sqrt{4+9}\) = \(\sqrt{13}\) units.