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Find the differential equation of all the circleswhich pass through the origin and whose centres lie on y-axis.A. `(x^(2)-y^(2))(dy)/(dx)-2xy=0`B. `(x^(2)-y^(2))(dy)/(dx)+2xy=0`C. `(x^(2)-y^(2))(dy)/(dx)-xy=0`D. `(x^(2)-y^(2))(dy)/(dx)+xy=0` |
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Answer» Correct Answer - A Equation of a circle is `x^(2)+(y-a)^(2)=a^(2)" "...(i)` `2x+2y(dy)/(dx)-2a(dy)/(dx)=0" "...(ii)` From Eqs. (i) and (ii) `(dy)/(dx)=(2xy)/(x^(2)-y^(2))` `implies(x^(2)-y^(2))(dy)/(dx)-2xy=0` |
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