Let C(x, y) be the centre of the circle passing through the points P(6, –6), Q(3, –7) and R(3, 3)
Then, PC = QC = RC(Being radius of the same circle)
PC2 = QC2 ⇒ (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2
⇒ x2 – 12x + 36 + y2 + 12y + 36 = x2 – 6x + 9 + y2 + 14y + 49
⇒ –12x + 12y + 6x – 14y + 72 – 58 = 0
⇒ – 6x – 2y + 14 = 0 ⇒ 3x + y –7 = 0 ...(i)
Also, QC2 = RC2
⇒ (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2
⇒ x2 – 6x + 9 + y2 + 14y + 49 = x2 – 6x + 9 + y2 – 6y + 9
⇒ 14y + 6y = 9 – 49 ⇒ 20y = – 40 ⇒ y = –2 ...(ii)
Putting y = –2 in (i), we get
3\(x\) + (–2) – 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3
∴ The centre is (3, –2).