# Ex 11.1, 10 - Chapter 11 Class 11 Conic Sections

Last updated at Sept. 28, 2018 by Teachoo

Last updated at Sept. 28, 2018 by Teachoo

Transcript

Ex 11.1, 10 Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16. Let the equation of the circle be (x – h)2 + (y – k)2 = r2. Since circle passes through point (4, 1) Point (4, 1) will satisfy the equation of circle Putting x = 4 & y = 1 in (A) (x – h)2 + (y – k)2 = r2. (4 − h)2 + (1 − k)2 = r2 (4)2 + h2 − 2(4)(h) + (1)2 + k2 − 2(1)(k) = r2 16 + h2 − 8h + 1 + k2 − 2k = r2 h2 + k2 − 8h − 2k + 17 = r2 Similarly, since circle passes through point (6, 5) Point (6, 5) will satisfy the equation of circle Putting x = 6, & y = 5 in (A) (x – h)2 + (y – k)2 = r2 (6 – h)2 + (5 – k)2 = r2 (6)2 + (h)2 − 2(6)(h) + (5)2 + (k)2 − 2(5)(k) = r2 36 + h2 − 12h + 25 + k2 − 10k = r2 h2 + k2 − 12h − 10k + 36 + 25 = r2 h2 + k2 − 12h − 10k + 61 = r2 Since centre of circle (h, k) lie on the line 4x + y = 16 i.e. centre (h, k) will satisfy the equation of line 4x + y = 16 Putting (h, k) in 4x + y = 16 4h + k = 16 Solving (1) & (2) h2 + k2 − 12h − 10k + 61 = r2 …(1) h2 + k2 − 8h − 2k + 17 = r2 …(2) Subtracting (2) from (1) (h2 + k2 − 8h − 2k + 17) − (h2 + k2 − 12h − 10k + 61)= r2 −r2 h2 + k2 − 8h − 2k + 17 − h2 − k2 + 12h + 10k − 61 = 0 h2 − h2 + k2 − k2 − 8h + 12h − 2k + 10k + 17 − 61 = 0 0 + 0 + 4h + 8k = 44 4h + 8k − 44 = 0 4h + 8k = 44 h + 2k = 11 Now our equations are 4h + k = 16 …(3) h + 2k = 11 …(4) From (4) h + 2k = 11 h = 11 – 2k Putting value of h in (3) 4(11 − 2k) + k = 16 44 − 8k + k = 16 44 − 7k = 16 −7k = 16 − 44 −7k = − 28 k = −28−7 k = 4 Putting value of k = 4 in (4) h + 2k = 11 h = 11 − 2(4) h = 11 − 8 h = 3 Hence, h = 3 & k = 3 Putting value of h, k in (1) h2 + k2 − 8h − 2k + 17 = r2 (3)2 + (4)2 − 8(3) − 2(4) + 17 = r2 9 + 16 − 24 − 8 + 17 = r2 10 = r2 r2 = 10 Equation of a circle is (x − h) 2 + (y − k) 2 = r2 Putting values (x − 4) 2 + (y − 3) 2 = 10 (x)2 + (4) 2 − 4(x)(4) + y2 + (3) 2 − 2(y)(3) = 10 x2 + 16 − 8x + y2 + 9 − 6y = 10 x2 + y2 − 8x − 6y + 16 + 9 − 10 = 0 x2 + y2 − 8x − 6y + 15 = 0 Which is the required equation of circle

Chapter 11 Class 11 Conic Sections

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.