Question

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

Answer 1

Let the radius of circle be a, then centre of the circle is (a,a). Hence, the equation of the circle is $(x-a{)}^2+(y-a{)}^2=a^2$ ...(i)
On differentiating Eq. (i) w.r.t. x, we get
$2x+2y{y}^{\prime }-2a-2a{y}^{\prime }=0\Rightarrow x+y{y}^{\prime }-a(1+y^{\prime })=0\Rightarrow a=\frac{x+y{y}^{\prime }}{1+y^{\prime }}$
On substituting this value of a in Eq. (i), we get
${(x-\frac{x+y{y}^{\prime }}{1+y^{\prime }})}^2+{(y-\frac{x+y{y}^{\prime }}{1+y^{\prime }})}^2={\left(\frac{x+y{y}^{\prime }}{1+y^{\prime }}\right)}^2\Rightarrow (x{y}^{\prime }-y{y}^{\prime }{)}^2+(y-x{)}^2=(x+y{y}^{\prime }{)}^2\Rightarrow (x-y{)}^2(y^{\prime }{)}^2+(x-y{)}^2=(x+y{y}^{\prime }{)}^2\Rightarrow (x-y{)}^2[1+(y^{\prime }{)}^2]=(x+y{y}^{\prime }{)}^2$
which is the required differential equation.