# Ex 11.1, 14 - Chapter 11 Class 11 Conic Sections

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 4−22+5 −22 = r r = 4−22+5 −22 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 = 132 (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 = 132 (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

Chapter 11 Class 11 Conic Sections

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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.