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Ex 7.4, 3 Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3). Let the circle pass through points Points A (6, −6), B (3, −7) and C (3, 3) Let O(x, y) be the centre of the circle Since the centre is equidistant from all the Points Lying on the circle. Distance AO = Distance CO Using distance formula, AO = √((𝑥−6)2+(𝑦+6)2) CO = √((𝑥−3)2+(𝑦−3)2) Now, AO = CO √((𝑥−6)2+(𝑦+6)2) = √((𝑥−3)2+(𝑦−3)2) Cancelling the square roots, (x − 6)2 + (y + 6)2 = (x − 3)2 + (y − 3)2 (x2 + 36 − 12x) − (x2 + 9 − 6x) = (y2 + 9 − 6y) − (y2 + 36 + 12y) Distance AO = Distance BO Using distance formula, AO = √((𝑥−6)2+(𝑦+6)2) BO = √((𝑥−3)2+(𝑦+7)2) Now, AO = BO √((𝑥−6)2+(𝑦+6)2) = √((𝑥−3)2+(𝑦+7)2) Cancelling the square roots (x − 6)2 + (y + 6)2 = (x − 3)2 + (y + 7)2 (x2 + 36 − 12x) − (x2 + 9 − 6x) = (y2 + 49 + 14y) − (y2 + 36 + 12y) x2 + 36 − 12x − x2 – 9 + 6x = y2 + 9 − 6y − y2 – 36 – 12y x2 − x2 − 12x + 6x + 36 − 9 = y2 – y2 − 6y − 12y + 9 − 36 −6x + 27 = −18y − 27 27 + 27 = −18y + 6x 54 = 6 (3y + x) 54/6 = −3y + x 9 = 3y + x x − 3y = 9 x2 + 36 − 12x − x2 − 9 + 6x = y2 + 49 + 14 − y2 − 36 − 12y x2 − x2 − 12x + 6x + 36 − 9 = y2 – y2 + 14y − 12y + 49 − 36 −6x + 27 = −2y + 13 27 − 13 = 2y + 6x 14 = 2 (y + 3x) 14/2 = y + 3x 7 = y + 3x 3x + y = 7 Hence, the equations are x − 3y = 9 3x + y = 7 From equation (1) x − 3y = 9 x = 9 + 3y Putting x = 9 + 3y in equation (2) 3x + y = 7 3 (9 + 3y) + y = 7 27 + 9y + y = 7 9y + y = 7 − 27 10y = −20 y = (−20)/10 y = −2 Putting y = –2 in equation (1) x − 3y = 9 x − 3 (−2) = 9 x + 6 = 9 x = 9 − 6 x = 3 Therefore, the center of the circle is = (x, y) = (3, −2).

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.