Last updated at Dec. 8, 2016 by Teachoo

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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 = −𝟓𝟐 Putting value of k = −52 in (3) h – 3k = 11 h = 11 + 3k h = 11 + 3(−52) h = 22 − 152 h = 𝟕 𝟐 Hence h = 72 & k = −52 Putting value of (h, k) = 72, −52 in (1) 2 – 722 + 3 – −522 = r2 4 − 722 + 6 + 522 = r2 −322 + 1122 = r2 94 + 1214 = r2 9 + 1214 = r2 1304 = r2 652 = r2 r2 = 𝟔𝟓𝟐 Now putting value of h, k & r2 in (A) (x − h)2 + (y − k)2 = r2 x − 722 + y − −522 = 652 x − 722 + y +522 = 652 x2 + 722 − 2(x) 72 + y2 + 254 + 5y = 652 x2 + 494 − 7x + y2 + 254 + 5y = 652 x2 + y2 − 7x + 5y + 494 + 254 = 652 x2 + y2 − 7x + 5y + 49 + 254 = 652 x2 + y2 − 7x + 5y + 744 = 652 x2 + y2 − 7x + 5y + 372 = 652 x2 + y2 − 7x + 5y = 652 − 372 x2 + y2 − 7x + 5y = 65 − 372 x2 + y2 − 7x + 5y = 282 x2 + y2 − 7x + 5y = 14 x2 + y2 − 7x + 5y − 14 = 0 Which is the required equation of circle

Chapter 11 Class 11 Conic Sections

Serial order wise

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .