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Ex 11.1, 12 - Find the equation of circle with radius 5 whose center

Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 4

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Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 5

Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 6
Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 7
Ex 11.1, 12 - Chapter 11 Class 11 Conic Sections - Part 8

 

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.