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Ex 11.1, 13 - Circle passing through (0, 0), making intercepts a

Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 3
Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 4
Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 5
Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 6
Ex 11.1,  13 - Chapter 11 Class 11 Conic Sections - Part 7

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Ex 11.1, 13 Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Let the equation of circle be (x – h)2 + (y – k)2 = r2 where (h, k) is the centre & r is the radius of a circle Also given that circle making intercepts a & b on the coordinate axes Let intercept on x-axis be a, So, coordinates of point A (a, 0) Let intercept on y-axis be b, So, coordinates of point B (0, b) Now, given that circle passes through (0, 0) So, Point (0, 0) will satisfy the equation of circle Putting x = 0 & y = 0 in (1) (0 − h)2 + (0 − k)2 = r2 h2 + k2 = r2 Also, point A (a, 0) lies on circle So, point A(a, 0) will satisfy the equation of circle Putting x = a & y = 0 in (1) (a − h)2 + (0 − k)2 = r2 (a − h)2 + k2 = r2 a2 + h2 − 2ah + k2 = r2 h2 + k2 + a2 − 2ah = r2 Putting h2 + k2 = r2 from (2) r2 + a2 − 2ah = r2 a2 − 2ah = r2 − r2 a(a − 2h) = 0 a − 2h = 0 h = (𝑎 )/2 Similarly, Point B(0, b) lie on the circle Point B(0, b) will satisfy the equation of circle Putting x = 0 & y = b in (1) (0 – h)2 + (b – k)2 = r2 h2 + (b – k)2 = r2 h2 + b2 + k2 − 2bk = r2 h2 + k2 + b2 − 2bk = r2 Putting h2 + k2 = r2 from (2) r2 + b2 − 2ah = r2 b2 − 2bk = r2 − r2 b(b − 2k) = 0 b − 2k = 0 b = 2k 2k = b k = (𝑏 )/2 Thus, h = (𝑎 )/2 & k = (𝑏 )/2 So, C(h, k) = (𝑎/2, 𝑏/2) Now we need to find radius of circle Putting h = 𝑎/2 & b = 𝑏/2 in (2) h2 + k2 = r2 (𝑎/2)^2 + (𝑏/2)^2 = r2 𝑎^2/4 + 𝑏^2/4 = r2 〖𝑎^2 + 𝑏〗^2/4 = r2 r2 = 〖𝑎^2 + 𝑏〗^2/4 Putting value of (h, k) & r2 in (1) (x – h)2 + (y – k)2 = r2 (𝑥− 𝑎/2)^2 + (𝑦−𝑏/2)^2 = 〖𝑎^2 + 𝑏〗^2/4 (x)2 + (𝑎/2)^2 − 2(x) (𝑎/2) + y2 +(𝑏/2)^2 − 2(y) (𝑏/2) = 〖𝑎^2 + 𝑏〗^2/4 x2 + 𝑎^2/4 − ax + y2 + 𝑏^2/4 − by = 〖𝑎^2 + 𝑏〗^2/4 x2 + y2 − ax − by + 𝑎^2/4 + 𝑏^2/4 = 〖𝑎^2 + 𝑏〗^2/4 x2 + y2 − ax − by + 〖𝑎^2 + 𝑏〗^2/4 = 〖𝑎^2 + 𝑏〗^2/4 x2 + y2 − ax − by = (𝑎2 + 𝑏2)/4 − 〖𝑎^2 + 𝑏〗^2/4 x2 + y2 − ax − by = 0 Which is required equation of circle

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Davneet Singh

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.