Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Options

A. `(x-y)^(2)[1+(y')]=[ x +yy']^(2)`
B. ` (x-y)^(2)[1+(y')^(2)] = (x+yy')^(2)`
C. ` (x+y)^(2)[1+(y')^(2)]= (x+yy')^(2)`
D. None of these

Answer

b

Explanation

Let the radius of a ciircle be a , then centre of the circle is (a,s)

Hence , The equation of family of circles is
`(x-a)^(2) + (y -a)^(2) = a^(2)" "` …(i)
` x^(2) +y^(2) - 2ax - 2ay +a^(2) =0`
On differenting Eq. (i) w.r.t x, we get
` 2x+ 2yy' - 2a - 2ay' = 0 `
` rArr 2x + 2yy' - 2a - 2ay' = 0 `
` rArr x + yy' - a ( 1+y') =0`
` rArr x + yy' - a (1+y') = 0`
` rArr a = (x + yy')/(1+y')` On substituting this value of in Eq,. (i) we get ` ( x - (x+yy')/(1+y'))^(2) + ( y - (x + yy')/(1+y'))^(2) = ((x+yy')/(1+y'))^(2)`
` rArr (xy' - yy')^(2) + ( y - x)^(2) = ( x+yy')^(2)`
` rArr (x-y)^(2) (y')^(2) + ( x-y)^(2) = ( x +yy')^(2)`
` rArr ( x =- y )^(2) [ 1 + (y')^(2) ] = ( x + yy')^(2)`
which is the required equation .