(a) (5, 2)Let A(8, 6) , B(8, –2) and C(2, –2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. Then, PA2 = PB2 = PC2 Now, PA2 = PB2 ⇒ (x – 8)2 + (y – 6)2 = (x – 8) + (y + 2)2 ⇒ x2 – 16x + 64 + y2 – 12y + 36 = x2 – 16x + 64 + y2 + 4y + 4 ⇒ 16y = 32 ⇒ y = 2. Now, PB2 = PC2 ⇒ (x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)2 ⇒ x2 – 16x + 64 + y2 + 4y + 4 = x2 – 4x + 4 + y2 + 4y + 4 ⇒ 12x = 60 ⇒ x = 5. ∴ Co-ordinates of the circumcentre are (5, 2).