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Differential Equation of a Curve from a Tangent | JEE

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Question
A tangent drawn at any point PP on a curve meets the xx-axis at QQ such that the circumcentre of △POQ\triangle POQ has abscissa half that of its ordinate. The differential equation of such a curve is:
Adydx=x+2y2x−y\dfrac{dy}{dx} = \dfrac{x + 2y}{2x - y}correct
Bdydx=2y−x2x+y\dfrac{dy}{dx} = \dfrac{2y - x}{2x + y}
Cdydx=2y−x2x+y\dfrac{dy}{dx} = \dfrac{2y - x}{2x + y}
DNone of these
Solution
Let P=(x,y)P = (x, y), O=(0,0)O = (0,0), and the tangent slope m=dydxm = \dfrac{dy}{dx}. The tangent meets the xx-axis at Q=(x−ym, 0)Q = \left(x - \dfrac{y}{m},\ 0\right). Step 1: Let the circumcentre be C=(t2, t)C = \left(\dfrac{t}{2},\ t\right) (abscissa half the ordinate). Since OO and QQ lie on the xx-axis, the perpendicular bisector of OQOQ is vertical, so the abscissa of CC is 12(x−ym)=t2\dfrac{1}{2}\left(x - \dfrac{y}{m}\right) = \dfrac{t}{2}, giving
t=x−ym.t = x - \frac{y}{m}.
 Step 2: Use CO=CPCO = CP:
t24+t2=(t2−x)2+(t−y)2⇒x2+y2=t(x+2y).\frac{t^2}{4} + t^2 = \left(\frac{t}{2} - x\right)^2 + (t - y)^2 \Rightarrow x^2 + y^2 = t(x + 2y).
 Step 3: Substitute t=x−ymt = x - \dfrac{y}{m} and simplify:
x2+y2=(x−ym)(x+2y)⇒y2−2xy=−y(x+2y)mx^2 + y^2 = \left(x - \frac{y}{m}\right)(x + 2y) \Rightarrow y^2 - 2xy = -\frac{y(x + 2y)}{m}
⇒x+2ym=2x−y⇒dydx=x+2y2x−y.\Rightarrow \frac{x + 2y}{m} = 2x - y \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{2x - y}.
 Correct answer: (1)
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