Question

The $D.E$ of the family of circles touching $X-axis$ at $(0,0)$ is
The $D.E$ of the family of circle passing through the origin and having their centre on $Y-axis$ is

• A. $\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$
• B. $\frac{dy}{dx}=\frac{x^2-y^2}{2xy}$
• C. $\frac{dy}{dx}=x^2-y^2$
• D. $\frac{dy}{dx}+\frac{x}{y}=0$

Answer 1

The correct option is A $\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$

The equation of circle is $(x-a{)}^2+(y-b{)}^2=r^2$

Center $=(a,b)$

Radius $=r$

Since, the circle touches the x-axis at origin then, center will be on the $y-axis$

So, $x-coordinate$ of center is $0.$

$i.e.a=0$

$\therefore$ $Center=(0,b)$

And, radius $=b$

So, equation of circle be

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

$\Rightarrow$ $x^2+(y-b{)}^2=b^2$

$\Rightarrow$ $x^2+y^2-2yb+b^2=b^2$

$\Rightarrow$ $x^2+y^2-2yb=0$

$\Rightarrow$ $x^2+y^2=2yb$ ------ ( 1 )

Differentiate w.r.t $x$

$\Rightarrow$ $2x+2y\frac{dy}{dx}=2b.\frac{dy}{dx}$

$\Rightarrow$ $x+y{y}^{\prime }=b.y^{\prime }$

$\Rightarrow$ $b=\left[\frac{x+y{y}^{\prime }}{y}\right]$

Putting value of $b$ in $\left(1\right)$

$\Rightarrow$ $x^2+y^2=2yb$

$\Rightarrow$ $x^2+y^2=2y\left[\frac{x+y{y}^{\prime }}{y^{\prime }}\right]$

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

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

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

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

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

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

$\Rightarrow$ $y^{\prime }=\frac{2xy}{x^2-y^2}$ $i.e.\frac{dy}{dx}=\frac{2xy}{x^2-y^2}$