Question

Form the differential equation of all circles which pass through origin and whose center lie on $\text{y-axis}$

Answer 1

Center of the circle lie on $\text{y-axis}$:
Let the center be $(0,k)$
Let the radius of the circle be $\text{r}$
Equation of the circle is
$(x-0{)}^2+(y-k{)}^2=r^2...\left(1\right)$
As the circle passing through origin,
so putting $(0,0)$ in $\left(1\right)$, we get
$\Rightarrow (0-0{)}^2+(0-k{)}^2=r^2$
$\Rightarrow r^2=k^2$
Therefore, the equation of circle becomes
$x^2+(y-k{)}^2=k^2$
$\Rightarrow x^2+y^2-2ky=0$
Differentiating w.r.t. $x$, we get
$2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0$
$\Rightarrow x+y\frac{dy}{dx}=k\frac{dy}{dx}$
$\Rightarrow x\frac{dx}{dy}+y=k$
Putting the value of $k$ in equation (2), we get
$x^2+y^2-2(x\frac{dx}{dy}+y)y=0$
$\Rightarrow x^2+y^2-2xy\frac{dx}{dy}-2y^2=0$
$\Rightarrow x^2-y^2-2xy\frac{dx}{dy}=0$
$\Rightarrow (x^2-y^2)\frac{dy}{dx}-2xy=0$
Hence, the required differential equation is
$(x^2-y^2)\frac{dy}{dx}-2xy=0$