The differential equation of all circles which pass through the origin and whose centers lie on the y-axis is
A
(x2−y2)dydx−2xy=0
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B
(x2−y2)dydx+2xy=0
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C
(x2−y2)dydx−xy=0
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D
(x2−y2)dydx+xy=0
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Solution
The correct option is A(x2−y2)dydx−2xy=0 Such family of circles is given by x2+(y−a)2=a2
or x2+y2−2ay=0
Differentiating, 2x+2ydydx=2adydx
or x+ydydx=adydx
Substituting the value of a in equation (1), we get (x2−y2)dydx=2xy (order in one again and degree 1)