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Ex 11.1, 11 - Circle passing through points (2, 3), (-1, 1) - Ex 11.1

Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 3 Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 4 Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 5 Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 6 Ex 11.1,  11 - Chapter 11 Class 11 Conic Sections - Part 7

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Ex 11.1, 11 Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0. Let the equation of the circle be (x – h)2 + (y – k)2 = r2. Since the circle passes through points (2, 3) Point (2, 3) will satisfy the equation of circle Putting x = 2, y = 3 in (A) (2 – h)2 + (3 – k)2 = r2 (2)2 + (h)2 − 2(2)(h) + (3)2 + k2 − 2(3)(k) = r2 4 + h2 − 4h + 9 + k2 − 6k = r2 h2 + k2 − 4h − 6k + 4 + 9 = r2 h2 + k2 − 4h − 6k + 13 = r2 Also, circle passes through the (−1, 1) Point (−1, 1) will satisfy the equation of circle Putting x = –1, y = –1 in (A) (−1 − h)2 + (1 − k)2 = r2 (−1)2 (1 + h) 2 + (1 − k) 2 = r2 (1 + h) 2 + (1 − k) 2 = r2 (1)2 + h2 + 2h + 1 + k2 − 2k = r2 1 + h2 + 2h + 1 + k2 − 2k = r2 h2 + k2 + 2h − 2k + 1 + 1 = r2 h2 + k2 + 2h − 2k + 2 = r2 Since centre (h, k) lie on the circle x − 3y − 11 = 0 Point (h, k) will satisfy the equation of line x − 3y − 11 = 0 So, h − 3k − 11 = 0 h − 3k = 11 Solving (1) & (2) h2 + k2 − 4h − 6k + 13 = r2 …(1) h2 + k2 + 2h − 2k + 2 = r2 …(2) Subtracting (1) from (2) (h2 + k2 − 4h − 6k + 13) − (h2 + k2+ 2h − 2k + 2) = r2 − r2 h2 + k2 − 4h − 6k + 13 − h2 − k2 − 2h + 2k − 2 = 0 h2 − h2 + k2 − k2 − 4h − 6k + 13 − 2h + 2k − 2 = 0 0 + 0 − 6h − 4k + 11 = 0 −6h − 4k = −11 6h + 4k = 11 Now our equations are h − 3k = 11 …(3) 6h + 4k = 11 …(4) From (3) h − 3k = 11 h = 11 + 3k Putting value of h = 11 + 3k in (4) 6h + 4k = 11 6(11 + 3k) + 4k = 11 66 + 18k + 4k = 11 22k = 11 − 66 k = ﷐−55 ﷮22﷯ k = ﷐−𝟓﷮𝟐﷯ Putting value of k = ﷐−5﷮2﷯ in (3) h – 3k = 11 h = 11 + 3k h = 11 + 3(﷐−5﷮2﷯) h = ﷐22 − 15﷮2﷯ h = ﷐𝟕 ﷮𝟐﷯ Hence h = ﷐7﷮2﷯ & k = ﷐−5﷮2﷯ Putting value of (h, k) = ﷐﷐7﷮2﷯, ﷐−5﷮2﷯﷯ in (1) ﷐﷐2 – ﷐7﷮2﷯﷯﷮2﷯ + ﷐﷐3 – ﷐−5﷮2﷯﷯﷮2﷯ = r2 ﷐﷐﷐4 − 7﷮2﷯﷯﷮2﷯ + ﷐﷐﷐6 + 5﷮2﷯﷯﷮2﷯ = r2 ﷐﷐﷐−3﷮2﷯﷯﷮2﷯ + ﷐﷐﷐11﷮2﷯﷯﷮2﷯ = r2 ﷐9﷮4﷯ + ﷐121﷮4﷯ = r2 ﷐9 + 121﷮4﷯ = r2 ﷐130﷮4﷯ = r2 ﷐65﷮2﷯ = r2 r2 = ﷐𝟔𝟓﷮𝟐﷯ Now putting value of h, k & r2 in (A) (x − h)2 + (y − k)2 = r2 ﷐﷐x − ﷐7﷮2﷯﷯﷮2﷯ + ﷐﷐y − ﷐﷐−5﷮2﷯﷯﷯﷮2﷯ = ﷐65﷮2﷯ ﷐﷐x − ﷐7﷮2﷯﷯﷮2﷯ + ﷐﷐y +﷐5﷮2﷯﷯﷮2﷯ = ﷐65﷮2﷯ x2 + ﷐﷐﷐7﷮2﷯﷯﷮2﷯ − 2(x) ﷐﷐7﷮2﷯﷯ + y2 + ﷐25﷮4﷯ + 5y = ﷐65﷮2﷯ x2 + ﷐49﷮4﷯ − 7x + y2 + ﷐25﷮4﷯ + 5y = ﷐65﷮2﷯ x2 + y2 − 7x + 5y + ﷐49﷮4﷯ + ﷐25﷮4﷯ = ﷐65﷮2﷯ x2 + y2 − 7x + 5y + ﷐49 + 25﷮4﷯ = ﷐65﷮2﷯ x2 + y2 − 7x + 5y + ﷐74﷮4﷯ = ﷐65﷮2﷯ x2 + y2 − 7x + 5y + ﷐37﷮2﷯ = ﷐65﷮2﷯ x2 + y2 − 7x + 5y = ﷐65﷮2﷯ − ﷐37﷮2﷯ x2 + y2 − 7x + 5y = ﷐65 − 37﷮2﷯ x2 + y2 − 7x + 5y = ﷐28﷮2﷯ x2 + y2 − 7x + 5y = 14 x2 + y2 − 7x + 5y − 14 = 0 Which is the required equation of circle

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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.