Question

The perpendicular from the origin to the tangent at any point on a curve is equal to the abscissa of the point of contact. Find the equation of the curve satisfying the above condition and which passes through $(1,1)$. (The curve is a circle with radius unity.)

Answer 1

Let the curve be $y=f\left(x\right)$.
Then the equation of the tangent at $(h,k)$ will be
$\frac{y-k}{x-h}=f^{\prime }(x{)}_{h,k}$
$y-k=f^{\prime }\left(x\right)(x-h)$
$f^{\prime }\left(x{)}_{h,k}\right(x-h)-(y-k)=0$
Distance from origin will be
$d=\frac{k-f^{\prime }(x{)}_{h,k}h}{\sqrt{f^{\prime }(x{)}^2+1}}$
Now $d=abscissa=h$
Hence
$h=\frac{k-f^{\prime }(x{)}_{h,k}h}{\sqrt{f^{\prime }(x{)}^2+1}}$
$h\sqrt{f^{\prime }(x{)}^2+1}+f^{\prime }\left(x\right)h=k$
$h[\sqrt{f^{\prime }(x{)}^2+1}+f^{\prime }(x\left)\right]=k$
$[\sqrt{f^{\prime }(x{)}^2+1}+f^{\prime }(x\left)\right]=\frac{k}{h}$
$\sqrt{f^{\prime }(x{)}^2+1}=\frac{k}{h}-f^{\prime }\left(x\right)$
Squaring both sides give us
$f^{\prime }(x{)}^2+1=\frac{k^2}{h^2}+f^{\prime }(x{)}^2-\frac{2k.f^{\prime }\left(x\right)}{h}$
$\frac{k^2}{h^2}-1=\frac{2k.f^{\prime }\left(x\right)}{h}$
Replacing $(h,k)$ by $(x,y)$ gives us
$\frac{y^2}{x^2}-1=\frac{2y}{x}\frac{dy}{dx}$
$\frac{y}{2x}-\frac{x}{2y}=\frac{dy}{dx}$
Let $y=vx$ Hence
$\frac{dy}{dx}=v+\frac{xdv}{dx}$
Hence
$\frac{v}{2}-\frac{1}{2v}=v+\frac{xdv}{dx}$
$\frac{v}{2}+\frac{1}{2v}=\frac{-xdv}{dx}$
$\frac{v^2+1}{2v}=\frac{-xdv}{dx}$
$\frac{dx}{-x}=\frac{2vdv}{v^2+1}$
$-lnx=ln(v^2+1)+lnC$
$lnx+ln(v^2+1)+lnC=0$
$Cx(v^2+1)=1$
$Cx(\frac{y^2}{x^2}+1)=1$
$Cx\left(\frac{y^2+x^2}{x^2}\right)=1$
$C{y}^2+C{x}^2=x$
It passes through $(1,1)$ hence
$2C=1$
Or
$C=\frac{1}{2}$
Hence the equation turns out to be
$x^2+y^2=2x$
Or
$(x-1{)}^2+y^2=1$