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Ex 11.1, 14 - Equation of a circle with centre (2, 2), passes - Ex 11.1

Ex 11.1,  14 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.1,  14 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.1,  14 - Chapter 11 Class 11 Conic Sections - Part 4

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Transcript

Ex 11.1, 14 (Method 1) Find the equation of a circle with centre (2, 2) and passes through the point (4, 5). We know that equation of circle is (x – h)2 + (y – k)2 = r2 Since, centre of circle is (2, 2) So h = 2 & k = 2 Our equation becomes (x – 2)2 + (y – 2)2 = r2 Now, Distance between centre and point on circle = radius Distance between points (h, 0) & (2,3) = r ﷐﷮﷐4−2﷯2+﷐5 −2﷯2﷯ = r r = ﷐﷮﷐4−2﷯2+﷐5 −2﷯2﷯ r = ﷐﷮22+32﷯ r = ﷐﷮4+9﷯ r = ﷐﷮13﷯ Putting value of r in (1) (x – 2)2 + (y – 2)2 = r2 (x – 2)2 + (y – 2)2 = ﷐﷐﷐﷮13﷯﷯﷮2﷯ (x – 2)2 + (y – 2)2 = 13 x2 + 22 – 4x + y2 + 22 – 4y = 13 x2 + 4 – 4x + y2 + 4 – 4y = 13 x2 + y2 – 4x – 4y = 13 – 4 – 4 x2 + y2 – 4x – 4y = 5 Hence, the required equation is x2 + y2 – 4x – 4y = 5 Ex 11.1, 14 (Method 2) Find the equation of a circle with centre (2, 2) and passes through the point (4, 5). We know that equation of circle is (x – h)2 + (y – k)2 = r2 Since, centre of circle is (2, 2) So h = 2 & k = 2 Our equation becomes (x – 2)2 + (y – 2)2 = r2 Since it is given that point (4, 5) passes through the circle Hence point (4, 5) will satisfy the equation of circle Putting x= h & y = 5 in (1) (x – 2)2 + (y – 2)2 = r2 (4 − 2)2 + (5 − 2)2 = r2 (2)2 + (3)2 = r2 4 + 9 = r2 13 = r2 r2 = 13 r = ﷐﷮13﷯ Putting value of r in (1) (x – 2)2 + (y – 2)2 = r2 (x – 2)2 + (y – 2)2 = ﷐﷐﷐﷮13﷯﷯﷮2﷯ (x – 2)2 + (y – 2)2 = 13 x2 + 22 – 4x + y2 + 22 – 4y = 13 x2 + 4 – 4x + y2 + 4 – 4y = 13 x2 + y2 – 4x – 4y = 13 – 4 – 4 x2 + y2 – 4x – 4y = 5 Hence, the required equation is x2 + y2 – 4x – 4y = 5

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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.