Ex 11.1, 10 - Chapter 11 Class 11 Conic Sections (Term 2)
Last updated at Feb. 6, 2020 by Teachoo
Ex 11.1
Last updated at Feb. 6, 2020 by Teachoo
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 (4, 1) It 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 42 + h2 − 8h + 12 + k2 − 2k = r2 16 + h2 − 8h + 1 + k2 − 2k = r2 h2 + k2 − 8h − 2k + 17 = r2 Since circle passes through (6, 5) It 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 62 + h2 − 12h + 52 + k2 − 10k = 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 = 0 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 = 4 Putting value of h, k in (1) h2 + k2 − 8h − 2k + 17 = r2 32 + 42 − 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 − 3) 2 + (y − 4) 2 = 10 x2 + 32 − 2(x)(3) + y2 + 42 − 2(y)(4) = 10 x2 + 9 − 6x + y2 + 16 − 8y = 10 x2 + y2 − 6x − 8y + 9 + 16 − 10 = 0 x2 + y2 − 6x − 8y + 15 = 0 Which is the required equation of circle