Find the differential equation of all circles with fixed radius and the centers are on the x-axis. Please 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 $\boxed{y^2((y')^2 + 1) = r^2}$

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MATHalino

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 $\boxed{y^2((y')^2 + 1) = r^2}$

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