Problem 55
Medium Separable and Homogeneous Equations

Find the curves for which the distance from the tangent line…

Find the curves for which the distance from the tangent line to the origin equals the $x$-coordinate of the point of tangency.

Explanation of Method

Write the point-to-line distance formula for the tangent at $(x,y)$, set it equal to $x$, square to clear the radical, and solve the resulting homogeneous ODE.

Step 1: Set up the equation

The tangent line $y'X-Y+(y-xy')=0$ has distance $\dfrac{|y-xy'|}{\sqrt{1+(y')^2}}$ from the origin. Setting this equal to $x$ and squaring,

\[(y-xy')^2=x^2[1+(y')^2] \quad\Longrightarrow\quad y^2-2xyy'=x^2 \quad\Longrightarrow\quad y'=\frac{y^2-x^2}{2xy}.\]

Step 2: Substitute $y=vx$

\[x\frac{dv}{dx}=\frac{v^2-1}{2v}-v=-\frac{v^2+1}{2v}.\]

Step 3: Separate and integrate

\[\frac{2v\,dv}{v^2+1}=-\frac{dx}{x} \quad\Longrightarrow\quad \ln(v^2+1)=-\ln|x|+C \quad\Longrightarrow\quad x(v^2+1)=K.\]

Step 4: Back-substitute $v=y/x$

\[\boxed{\,x^2+y^2=Cx\ \ \Longleftrightarrow\ \ \left(x-\tfrac{C}{2}\right)^2+y^2=\tfrac{C^2}{4}}\]

— a family of circles through the origin, centered on the $x$-axis.


Result

Final answer:

\[\boxed{\,x^2+y^2=Cx.}\]