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

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is :

(x -a)2 + (y –a)2 = a2 -----------(1)

Now differentiating above equation w.r.t. x, we get,

2(x-a) + 2(y-a) = 0

(x – a) + (y – a)y’ = 0

x – a +yy’ – ay’ = 0

x + yy’ –a(1+y’) = 0

a =

Now, substituting the value of a in equation (1), we get,

(x - y)2.y’2 + (x – y)2 = (x + yy’)2

(x – y)2[1 + (y’)2] = (x + yy’)2

Therefore, the required differential equation of the family of circles is

(x – y)2[1 + (y’)2] = (x + yy’)2

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz on DIfferential Calculus50 mins
Functional Equations - JEE with ease48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses