Differential Equations (Families of Curves): Circles with fixed radius r and tangent to the y-axis.

The general equation of a circle with centre at (h, k) and radius r is given by
${\displaystyle {(x-h)}^2+{(y-k)}^2=r^2}$
Also, it is given that y axis is tangent to this circle.
This means the perpendicular distance from the centre of the circle to the line ${\displaystyle x=0}$ is equal to the radius of the circle.
This means ${\displaystyle h=r}$
Hence, the equation of such a circle becomes
${\displaystyle {(x-r)}^2+{(y-k)}^2=r^2}$
Here, k is hte only variable left.
Hence, to obtain the differential equation, we need to eliminate the variable k
We will do this with the help of differentiation.
Differentiating the curve with respect to x, we get
${\displaystyle 2(x-r)+2(y-k)\frac{dy}{dx}=0}$
${\displaystyle (y-k)\frac{dy}{dx}=r-x}$
${\displaystyle (y-k)=\frac{r-x}{\frac{dy}{dx}}}$
Putting the value of ${\displaystyle y-k}$ back in the equation we get
${\displaystyle {(x-r)}^2+{(y-k)}^2=r^2}$
${\displaystyle {(x-r)}^2+\left(\frac{r-x}{\frac{dy}{dx}}\right)=r^2}$
${\displaystyle {\left(\frac{dy}{dx}\right)}^2{(x-r)}^2+{(x-r)}^2={\left(\frac{dy}{dx}\right)}^2\left(r^2\right)}$
${\displaystyle {\left(\frac{dy}{dx}\right)}^2(x^2+r^2-2xr-r^2)+{(x-r)}^2=0}$
${\displaystyle {\left(\frac{dy}{dx}\right)}^2(x^2-2xr)+{(x-r)}^2=0}$
${\displaystyle {\left(\frac{dy}{dx}\right)}^2=\frac{{(x-r)}^2}{2xr-x^2}}$

${\displaystyle \Rightarrow {(x-a)}^2+y^2=r^2}$
${\displaystyle \Rightarrow 2(x-a)+2y{y}_1=0}$
${\displaystyle \Rightarrow (x-a)=-y{y}_1}$
$\Rightarrow y^2{y}_1^2+y^2=r^2$

Center is (0,k) and radius is r.
Equation is
$x^2+(y-k{)}^2=r^2$ Equation 1
By differentiating we get,
$2x+2(y-k)\frac{dy}{dx}=0$
$k=+x\frac{dx}{dy}+y$ Equation 2
Putting the value of K from Equation 1 and 2, we get
$\begin{array}{}{x}^2+(y-x\frac{dx}{dy}-y{)}^2=r^2\\ x^2+x^2(\frac{dx}{dy}{)}^2=r^2\\ 1+(\frac{dx}{dy}{)}^2=\frac{r^2}{x^2}\\ (\frac{dy}{dx}{)}^2=\frac{x^2}{r^2-x^2}\end{array}$