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Question

The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is

A
(x2y2)dydx2xy=0
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B
(x2y2)dydx+2xy=0
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C
(x2y2)dydxxy=0
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D
(x2y2)dydx+xy=0
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Solution

The correct option is A (x2y2)dydx2xy=0
Such family of circles is given by
x2+(ya)2=a2
or x2+y22ay=0
Differentiating, 2x+2ydydx=2adydx
or x+ydydx=adydx
Substituting the value of a in equation (1), we get
(x2y2)dydx=2xy (order in one again and degree 1)

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