Differential Equations (PLEASE HELP PO)

The center of the circle is $(a, 0)$ and the fixed radius is, say, $r$.

Thus the equation of the circle is $(x-a)^2+y^2 = r^2$. Since $a$ is the arbitrary constant here, get the derivative once.

\begin{eqnarray*}
(x-a)^2+y^2 &=& r^2\\
2(x-a) + 2y y' &=& 0\\
(x-a) + yy' &=& 0\\
x-a &=& -yy'
\end{eqnarray*}
Then substitute to the original equation.
\begin{eqnarray*}
(x-a)^2 + y^2 &=& r^2\\
(yy')^2 + y^2 &=& r^2\\
y^2((y')^2 + 1) &=& r^2
\end{eqnarray*}

Thus the differential equation is $\boxed{y^2((y')^2 + 1) = r^2}$