      1. Chapter 11 Class 11 Conic Sections
2. Serial order wise
3. Ex 11.1

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 2 1 2+ 2 1 2 = 5 2 2+ 3 0 2 = 5 2 2+ 3 2 = 5 2 2+ 2 2 2 +9 = 5 4+ 2 4 +9 = 5 13+ 2 4 = 5 Squaring both sides ( 13+ 2 4 )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

Ex 11.1 