Find the family of curves for which the length of the…

Find the family of curves for which the length of the tangent (from the point of tangency to where it meets the $x$-axis) equals the segment the tangent cuts off on the positive $x$-axis.

Step 1: Set up the equation

The tangent meets the $x$-axis at $\left(x-\dfrac{y}{y'},0\right)$, so the $x$-intercept (distance from the origin) is $\left|x-\dfrac{y}{y'}\right|$, and the tangent length is $\dfrac{|y|\sqrt{1+(y')^2}}{|y'|}$. Setting these equal and simplifying (multiplying through by $y'$) gives $y-xy'=\pm y\sqrt{1+(y')^2}$; squaring,

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

Step 2: Substitute $y=vx$

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

Step 3: Separate and integrate

Using $\dfrac{1-v^2}{v(1+v^2)}=\dfrac1v-\dfrac{2v}{1+v^2}$,

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

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

\[\frac{y/x}{1+y^2/x^2}=Kx \quad\Longrightarrow\quad \frac{xy}{x^2+y^2}=Kx \quad\Longrightarrow\quad \frac{y}{x^2+y^2}=K.\]

Equivalently, letting $C=1/K$,

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

— circles through the origin, centered on the $y$-axis and tangent there to the $x$-axis.