IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

sumit_grg

the differential eq. of the system of circle touching the x- axis at origin is

prash_shan_jpr

Assume d center to b (0,a)

after wards

use x^2 + (y-a)^2 = a^2

open d bracket

u will get x^2 + y^2 = 2ay

differentiate it once

then put d value of a.

joyfrancis

the circle touches the x axis at the orgin so the center must lie on the y axis..

let center be (0,a)..

now radius is also equal to 'a' in this case..

ther4 eqn of circle is ..

x² + (y-a)² = a²

=> a = (x² + y²) / 2y

differentiating the eqn of circle once only since we have only one parameter "a".

we have

dy/dx = x/a-y

substitute value of a ..

answer is .

2xy/(x² - y²) = dy/dx

mathwiz

the eqn for circles touching x axis at origin is x^2+y^2-2ky=0--(1)
differentiate this eqn with res to x and find the value of k
substitute its value in the first eqn..(1)
u'll get the ans
x^2-y^2-2xy/(dy/dx)=0

please correct me if i am wrong..

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