Engineering Board Exam Reviewer · Orthogonal Trajectories · hard

Find the orthogonal trajectories of the family of circles x2+y2=2Cx.

  1. x2 + y2 = K x
  2. x2 - y2 = K y
  3. x2 + y2 = K
  4. x2 + y2 = K y

Answer

D. x2 + y2 = K y

Solution

Differentiate x2+y2=2Cx: x+yy'=C=x2+y2/2x y'=y2-x2/2xy. Orthogonal family: y'=2xy/x2-y2, which is homogeneous. Let y=vx: xv'=2v/1-v2-v=v(1+v2)/1-v2, so ≤ft[ 1v-2v/1+v2 ]dv=dx/x /1+v2= x+c. Then xy/x2+y2=Kx, i.e. x2+y2=K'y (circles tangent to the x-axis).

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