web analytics

Ex 11.1, 3 - Equation of circle center (1/2, 1/4) radius 1/12   - Circle

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

Transcript

Ex 11.1, 3 Find the equation of the circle with centre ﷐﷐1﷮2﷯,﷐1﷮4﷯﷯ and radius ﷐1﷮12﷯ We know that equation of a circle is (x­ – h)2 + (y ­– k)2 = r2 Where (h, k) is the centre & r is the radius Here Centre (h, k) = ﷐﷐1﷮2﷯, ﷐1﷮4﷯﷯ So h = ﷐1﷮2﷯ & k = ﷐1﷮4﷯ & radius (r) = ﷐1﷮12﷯. Putting values in (1) ﷐﷐x − ﷐1﷮2﷯﷯﷮2﷯ + ﷐﷐y – ﷐1﷮4﷯﷯﷮2﷯ = ﷐﷐﷐1﷮12﷯﷯﷮2﷯ x2 + ﷐﷐﷐1﷮2﷯﷯﷮2﷯ − 2 × x × ﷐1﷮2﷯ + y2 + ﷐﷐﷐1﷮4﷯﷯﷮2﷯ − 2 × y × ﷐1﷮4﷯ = ﷐1﷮144﷯ x2 + ﷐1﷮4﷯ − x + y2 + ﷐1﷮16﷯ − ﷐1﷮2﷯y = ﷐1﷮144﷯ x2 + y2 − x − ﷐1﷮2﷯y + ﷐1﷮4﷯ + ﷐1﷮16﷯ − ﷐1﷮144﷯= 0 ﷐144𝑥2 + 144𝑦2 − 144𝑥 − 72𝑦 + 34 + 9 + 1﷮144﷯ = 0 144x2 + 144y2 − 144x − 72y + 44 = 0 × 144 144x2 + 144y2 − 144x − 72y + 44 = 0 4(36x2 + 36y2 − 36x − 18y + 11) = 0 36x2 + 36y2 − 36x − 18y + 11 = 0 Hence the required equation of circle is 36x2 + 36y2 − 36x − 18y + 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 provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail