Orthogonal Trajectroes of System of Circles Centred on x Axis Passing Through the Origin

Consider the set of circles passing through the origin, with centres on the  
\[x\]
  axis. The equations have equations of the form  
\[(x-a)^2+y^2=a^2\]
  where  
\[a\]
  is the radius.
Differentiating and simplifying gives  
\[x-a+y \frac{dy}{dx}=0 \rightarrow a=x+ y \frac{dy}{dx}\]
.
Substitute this expression for  
\[a\]
  into the equation of the circle and simplify to give  
\[x^2+y^2=2x(x+y \frac{dy}{dx} ) \rightarrow \frac{dy}{dx} = \frac{y^2-x^2}{2xy}\]
. The perpendiculars to the circle at every point on the circle have gradient  
\[-1/ \frac{dy}{dx} = \frac{2xy}{x^2-y^2}\]
.
This perpendicular gradient function can be written as  
\[\frac{dx}{dy}=\frac{x^2-y^2}{2xy}\]
. This equation is homogeneous so substitute  
\[x=vy \rightarrow \frac{dx}{dy}=v+y \frac{dv}{dy} \rightarrow v+ y \frac{dv}{dy}= \frac{y^2v^2-y^2}{2vyy}= \frac{v^2-1}{2v} \]
.
A little rearrangement followed by separation of variables gives  
\[\frac{2v}{v^2+1} dv + \frac{1}{y} dy\]
.
Integration gives  
\[ln(1+v^2)+ln(y)=C \rightarrow ln((1+x^2/y^2)y)=C \rightarrow x^2+y^2=by\]
  where  
\[b=e^C\]
.
We can rearrange this equation as  
\[x^2+(y-b/2)^2=b^2/4\]
  which is the equation of a circle on with centre at  
\[y=b/2\]
, radius  
\[b/2\]
.

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