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Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).

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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).

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