Answer :

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


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