1.

The differential equation representing all possible curves that cut each member of the family of circles `x^(2)+y^(2)-2Cx=0` (C is a parameter) at right angle, isA. `(dy)/(dx)=(2xy)/(x^(2)+y^(2))`B. `(dy)/(dx)=(2xy)/(x^(2)-y^(2))`C. `(dy)/(dx)=(x^(2)+y^(2))/(2xy)`D. `(dy)/(dx)=(x^(2)-y^(2))/(2xy)`

Answer» Correct Answer - B
Here, we have to find the orthogonal trajectories of the family of circles
`x^(2)+y^(2)-2Cx=0" …(i)"`
Differentiating (i) w.r.t. x, we get
`2x+2y(dy)/(dx)-2C=0rArrC=x+y(dy)/(dx)" ...(ii)"`
From (i) and (ii), we obtain
`x^(2)+y^(2)-2x(x+y(dy)/(dx))=0" [By eliminating C]"`
`rArr" "y^(2)-x^(2)-2xy(dy)/(dx)=0rArry^(2)-x^(2)=2xy(dy)/(dx)" ...(iii)"`
This is the differential equation representing the given family of circles.
To find the differential equation of the orthogonal trajectories,
we replace `(dy)/(dx)by-(dx)/(dy)` in equation (iii).
Thus, the differential equation representing the orthogonal trajectories is
`y^(2)-x^(2)=-2xy(dx)/(dy)or, (dy)/(dx)=-(2xy)/(x^(2)-y^(2))`


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