# Example 4 - Chapter 11 Class 11 Conic Sections

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 4 Find the equation of the circle which passes through the points (2, 2), and (3,4) and whose centre lies on the line x + y = 2. Equation of circle with centre (h, k) is (x h)2 + (y k)2 = r2. Since the circle passes through (2, 2) Point (2, 2) will satisfy the equation of circle Putting x = 2, y = 2 in (A) (2 h)2 + ( 2 k)2 = r2 4 + h2 4h + 4 + ( 2)2 + ( k)2 + 2(-2)(k) = r2 4 + h2 4h + 4 + k2 + 4k = r2 h2 + k2 4h + 4k + 8 = r2 Similarly , since circle passes through (3, 4) Point (3, 4) will satisfy the equation of circle Putting x = 3, y = 4 in (A) (3 h)2 + (4 k)2 = r2 (3)2 + (h)2 2(3)(h) + (4)2 + (k)2 2(4)(k) = r2 9 + h2 6h + 16 + k2 8k = r2 h2 + k2 6h 8k + 25 = r2 Subtracting (2) from (1) (h2 + k2 6h 8k + 25) (h2 + k2 4h + 4k + 8) = r2 r2 h2 + k2 4h + 4k + 8 h2 k2 + 6h + 8k 25 = 0 h2 h2 + k2 k2 4h + 6h + 4k + 8k + 8 25 = 0 0 + 0 + 2h + 12k 17 = 0 2h + 12k 17 = 0 Since center (h, k) lie on the line x + y = 2 i.e. point (h, k) will satisfy the equation of circle h + k = 2 Now our equations are 2h + 12k = 17 (3) h + k = 2 (4) From (4) h + k = 2 k = 2 h Putting value of k in (3) 2h + 12(2 h) = 17 2h + 24 12h = 17 10h = 17 24 10h = 7 h = 7 10 h = 0.7 Putting value of h = 0.7 in k = 2 h k = 2 (0.7) k = 1.3 Putting value of (h, k) = (0.7, 1.3) in (A) (x h)2 + (y k)2 = r2 (2 0.7)2 + ( 2 1.3)2 = r2 (1.3)2 + (3.3)2 = r2 1.69 + 10.89 = r2 12.58 = r2 r2 = 12.58 Putting value of h, k & r2 in (A) (x 0.7)2 + (y 1.3)2 = 12.58 Hence required equation circle is (x 0.7)2 + (y 1.3)2 = 12.58

Chapter 11 Class 11 Conic Sections

Serial order wise

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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.