Find the equation of the family of all erhogonal trajectories of the family ofcircles which pass through origin and hav centres on the y-axis.
Upon differentiation with respect to x, we obtain
2x+2yy′+2g(1−y′)=0,
where y′ denotes the derivative of y with respect to x. Upon dividing out by 2, we arrive at
x+yy′+g(1−y′)=0,
from which we get
g=x+yy′y′−1.
Putting this value of g into the equation of the family of circles, we get
x2+y2+2x+yy′y′−1(x−y)−2=0,
so
(x2+y2−2)(y′−1)+2(x+yy′)(x−y)=0
or
(x2+y2−2+2xy−2y2)y′+(2x2−2xy−x2−y2+2)=0
or
(x2+2xy−y2−2)y′+(x2−2xy−y2+2)=0,
from which we get
y′=−
Now for the orthogonal trajectories, we get
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