Last updated at Dec. 8, 2016 by Teachoo

Transcript

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) = 𝑎2, 𝑏2 Now we need to find radius of circle Putting h = 𝑎2 & b = 𝑏2 in (2) h2 + k2 = r2 𝑎22 + 𝑏22 = r2 𝑎24 + 𝑏24 = r2 𝑎2 + 𝑏24 = r2 r2 = 𝑎2 + 𝑏24 Putting value of (h, k) & r2 in (1) (x – h)2 + (y – k)2 = r2 𝑥− 𝑎22 + 𝑦−𝑏22 = 𝑎2 + 𝑏24 (x)2 + 𝑎22 − 2(x) 𝑎2 + y2 +𝑏22 − 2(y) 𝑏2 = 𝑎2 + 𝑏24 x2 + 𝑎24 − ax + y2 + 𝑏24 − by = 𝑎2 + 𝑏24 x2 + y2 − ax − by + 𝑎24 + 𝑏24 = 𝑎2 + 𝑏24 x2 + y2 − ax − by + 𝑎2 + 𝑏24 = 𝑎2 + 𝑏24 x2 + y2 − ax − by = 𝑎2 + 𝑏24 − 𝑎2 + 𝑏24 x2 + y2 − ax − by = 0 Which is required equation of circle

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.