The differential equation for all family of lines which are at a unit distance from the origin is:

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${(y - x \frac{dy}{dx})}^{2} = 1 + {(\frac{dy}{dx})}^{2}$

${(y - x \frac{dy}{dx})}^{2} = 1 - {(\frac{dy}{dx})}^{2}$

${(y + x \frac{dy}{dx})}^{2} = 1 + {(\frac{dy}{dx})}^{2}$

${(y + x \frac{dy}{dx})}^{2} = 1 - {(\frac{dy}{dx})}^{2}$

answer is C.

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Detailed Solution

The equation of a straight line which is at a unit distance from the origin is

$xcos θ + ysin θ = 1 \to (1)$

Differentiating w.r.t. x

$\cos α + \frac{dy}{dx} \sin α = 0 \to (2)$

On eliminating α from (1) and (2), we get

$\begin{array}{l} \sin α (y - x \frac{dy}{dx}) = 1 \\ \Rightarrow (y - x \frac{dy}{dx}) = cosec α \to (3) \end{array}$

Also, Slope $= \frac{dy}{dx} = - \cot α$ {using(2}

$cosec α = \sqrt{1 - \cot^{2} α} = \sqrt{1 - {(\frac{dy}{dx})}^{2}} \to (4)$

From (3) and (4), we have ${(y - \frac{xdy}{dx})}^{2} = \{1 + {(\frac{dy}{dx})}^{2}\}$

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