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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.