Question

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

Answer 1

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)

Differentiating equation(1)with respect to x,we get:

\(2(x-α)+2(y-α)\frac{dy}{dx}=0\)

⇒(x-α)+(y-α)y=0

⇒x-α+yy-αy=0

⇒x+yy-α(1+y)=0

\(⇒α=\frac{x+yy}{1+y}\)

Substituting the value of a in equation(1),we get:

\([x-(\frac{x+yy}{1+y})]^2+[y-(\frac{x+yy}{1+y})]^2=(\frac{x+yy}{1+y})^2\)

\(⇒[(\frac{x-y)y}{(1+y}]^2+[\frac{y-x}{1+y}]^2=[\frac{x+yy}{1+y'}]^2\)

\(⇒(x-y)^2.y^2+(x-y)^2=(x+yy)^2\)

\(⇒(x-y)^2[1+(y)^2]=(x+yy)^2\)

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

⇒(x-y)^2[1+(y)^2]=(x+yy)^2