Question

Find the differential equation of the family of all the circles.
(A) touching X-axis at the origin.
(B) touching Y-axis at the origin.

Answer 1

• (A) Family of all the circles touching $x$-axis at origin

We know that, equation of circle is-

${(x-h)}^2+{(y-k)}^2=r^2$

Whereas,

Centre of circle $\equiv (h,k)$

Radius of circle $=r$

Since the circle touches the $x$-axis at origin, thus centre will be on $y$-axis.

$\therefore h=0$

$\therefore$ Centre of the circle $\equiv (0,k)$

Thus, radius of the circle $=k$

Now, equation of the circle-

${(x-0)}^2+{(y-k)}^2=k^2$

$x^2+y^2+k^2-2yk=k^2$

$x^2+y^2-2yk=0.....\left(1\right)$

As the above equation has only one variable, thus differentiating the above equation once w.r.t. $x$, we have

$2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0$

$x+y{y}^{\prime }-k{y}^{\prime }=0$

$\Rightarrow k=\frac{x+y{y}^{\prime }}{y^{\prime }}$

Substituting the value of $k$ in ${eq}^n\left(1\right)$, we have

$x^2+y^2-2y\left(\frac{x+y{y}^{\prime }}{y^{\prime }}\right)=0$

$(x^2+y^2)y^{\prime }=2y(x+y{y}^{\prime })$

$x^2{y}^{\prime }+y^2{y}^{\prime }=2xy+2y^2{y}^{\prime }$

$\Rightarrow x^2{y}^{\prime }+y^2{y}^{\prime }-2y^2{y}^{\prime }=2xy$

$\Rightarrow y^{\prime }(x^2-y^2)=2xy$

$\Rightarrow y^{\prime }=\frac{2xy}{(x^2-y^2)}$

• (B) Family of all the circles touching $y$-axis at origin

We know that, equation of circle is-

${(x-h)}^2+{(y-k)}^2=r^2$

Whereas,

Centre of circle $\equiv (h,k)$

Radius of circle $=r$

Since the circle touches the $y$-axis at origin, thus centre will be on $x$-axis.

$\therefore k=0$

$\therefore$ Centre of the circle $\equiv (h,0)$

Thus, radius of the circle $=h$

Now, equation of the circle-

${(x-h)}^2+{(y-0)}^2=k^2$

$x^2+h^2-2xh+y^2=h^2$

$x^2+y^2-2hx=0.....\left(1\right)$

As the above equation has only one variable, thus differentiating the above equation once w.r.t. $x$, we have

$2x+2y\frac{dy}{dx}-2h=0$

$x+y{y}^{\prime }-h=0$

$\Rightarrow h=x+y{y}^{\prime }$

Substituting the value of $h$ in ${eq}^n\left(1\right)$, we have

$x^2+y^2-2x(x+y{y}^{\prime })=0$

$(x^2+y^2)=2x(x+y{y}^{\prime })$

$x^2+y^2=2x^2+2xy{y}^{\prime }$

$\Rightarrow x^2+y^2-2x^2=2xy{y}^{\prime }$

$\Rightarrow (y^2-x^2)=2xy{y}^{\prime }$

$\Rightarrow y^{\prime }=\frac{(y^2-x^2)}{2xy}$

Hence family of all the circles touching $x$-axis at origin is $y^{\prime }=\frac{2xy}{(x^2-y^2)}$ while family of all the circles touching $y$-axis at origin is $y^{\prime }=\frac{(y^2-x^2)}{2xy}$.