Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide31.JPG

Slide32.JPG
Slide33.JPG Slide34.JPG

Slide35.JPG Slide36.JPG Slide37.JPG Slide38.JPG

 

  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise

Transcript

Ex 11.1, 12 (Method 1) Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3). We know that equation of circle is (x – h)2 + (y – k)2 = r2 Centre of circle is denoted by (h, k) Since it lies on x-axis , k = 0 Hence Centre of circle = (h, 0) & given Radius = 5 Distance between centre and point on circle = radius Distance between points (h, 0) & (2,3) = 5 Distance between points (h, 0) & (2,3) = 5 √((𝑥2−𝑥1)2+(𝑦2−𝑦1)2) = 5 √((2 −ℎ)2+(3−0)2) = 5 √((2 −ℎ)2+3^2 ) = 5 √((2)2+(ℎ)2−2(2)ℎ+9) = 5 √(4+ℎ2−4ℎ+9) = 5 √(13+ℎ2−4ℎ) = 5 Squaring both sides (√(13+ℎ2 −4ℎ))2 = 52 13 + h2 − 4h = 25 h2 – 4h + 13 − 25 = 0 h2 − 4h − 12 = 0 h2 − 6h + 2h − 12 = 0 h(h − 6) + 2(h − 6) = 0 (h + 2)(h − 6) = 0 So, h = –2 & h = 6 When h = –2 Equation of circle is (x – h)2 + (y – k)2 = r2 (x − (−2)2) + (y – 0)2 = 52 (x + 2)2 + y2 = 25 (x)2 + (2)2 + 2(x)(2) + y2 = 25 x2 + y2 + 4x + 4 − 25 = 0 x2 + y2 + 4x − 21 = 0 When h = 6 Equation of circle is (x – h)2 + (y – k)2 = r2 (x − 6)2 + (y – 0)2 = 52 x2 + (6)2 − 2(x)(6) + y2 = 25 x2 + 36 − 12x + y2 = 25 x2 + y2 − 12x + 36 − 25 = 0 x2 + y2 − 12x + 11 = 0 Hence required equation of circle is x2 + y2 + 4x − 21 = 0 or x2 + y2 − 12x + 11 = 0 Ex 11.1, 12 (Method 2) Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3). We know that equation of circle is (x – h)2 + (y – k)2 = r2 Centre of circle is denoted by (h, k) Since it lies on x-axis , k = 0 Hence centre of circle = (h, 0) & given radius = 5 So, the equation of circle is (x − h)2 + (y − 0)2 = (5)2 (x − h)2 + y2 = 25 Given that Circle passes through the point (2, 3) So, point (2, 3) will satisfy the equation of circle Putting x = 2 & y = 3 in (1) (2 − h) 2 + (3) 2 = 25 (2) 2 + (h) 2 − 2(2)(h) + 9 = 25 4 + h2 − 4h + 9 = 25 h2 − 4h + 9 + 4 − 25 = 0 h2 − 4h − 12 = 0 h2 − 6h + 2h − 12 = 0 h(h − 6) + 2(h − 6) = 0 (h + 2)(h − 6) = 0 So, h = –2 & h = 6 When h = –2 Equation of circle is (x – h)2 + (y – k)2 = r2 (x − (−2)2) + (y – 0)2 = 52 (x + 2)2 + y2 = 25 (x)2 + (2)2 + 2(x)(2) + y2 = 25 x2 + y2 + 4x + 4 − 25 = 0 x2 + y2 + 4x − 21 = 0 When h = 6 Equation of circle is (x – h)2 + (y – k)2 = r2 (x − 6)2 + (y – 0)2 = 52 x2 + (6)2 − 2(x)(6) + y2 = 25 x2 + 36 − 12x + y2 = 25 x2 + y2 − 12x + 36 − 25 = 0 x2 + y2 − 12x + 11 = 0 Hence required equation of circle is x2 + y2 + 4x − 21 = 0 or x2 + y2 − 12x + 11 = 0

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.