Q60 / 120 · VITEEE 2008 · Mathematics

The differential equation of all straight lines touching the circle $x^{2}+y^{2}=a^{2}$ is

  • A
    $\left(\mathrm{y}-\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=\mathrm{a}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]$
  • Correct answer
    B
    $\left(\mathrm{y}-\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}=\mathrm{a}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]$
  • C
    $\left(y-x \frac{d y}{d x}\right)=a^{2}\left[1+\left(\frac{d y}{d x}\right)\right]$
  • D
    $\left(y-\frac{d y}{d x}\right)=a^{2}\left[1-\frac{d y}{d x}\right]$

Solution

The given circle is, $x^{2}+y^{2}=a^{2}$ Differentiating with respect to $x$, we get $\begin{array}{l} 2 x+2 y \frac{d y}{d x}=0 \Rightarrow \quad x+y \frac{d y}{d x}=0 \\ \Rightarrow\left(x+y \frac{d y}{d x}\right)^{2}=0 \quad \text { (Squaring both sides) } \end{array}$ $\Rightarrow x^{2}+2 x y \frac{d y}{d x}+y^{2}\left(\frac{d y}{d x}\right)^{2}=0$ $\Rightarrow -2 x y \frac{d y}{d x}=x^{2}+y^{2}\left(\frac{d y}{d x}\right)^{2}$ $\Rightarrow y^{2}+x^{2}\left(\frac{d y}{d x}\right)^{2}-2 x y \frac{d y}{d x}$ $= x^{2}+y^{2}+x^{2}\left(\frac{d y}{d x}\right)^{2}+y^{2}\left(\frac{d y}{d x}\right)^{2}$ $\Rightarrow\left(\right.$ Adding $y^{2}+x^{2}\left(\frac{d y}{d x}\right)^{2}$ both sides $)$ $\begin{aligned} \Rightarrow\left(y-x \frac{d y}{d x}\right)^{2}=a^{2}\left[1+\left(\frac{d y}{d x}\right)^{2}\right] & \\ &\left(\because x^{2}+y^{2}=a^{2}\right) \end{aligned}$

Hence, the correct answer is option: 'B'.

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