Last updated at May 29, 2018 by Teachoo

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−𝑥12+𝑦2−𝑦12 = 5 2 −ℎ2+3−02 = 5 2 −ℎ2+32 = 5 22+ℎ2−22ℎ+9 = 5 4+ℎ2−4ℎ+9 = 5 13+ℎ2−4ℎ = 5 Squaring both sides (13+ℎ2 −4ℎ)2 = 25 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 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

Chapter 11 Class 11 Conic Sections

Serial order wise

About the Author

Davneet Singh

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.