Question

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

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

Answer 1

Answer: (B)

The given circle is, x$^2$ + y$^2$ = a$^2$
Differentiating with respect to x, we get
$2x + 2y\frac{dy}{dx}=0 \Rightarrow \quad x+y\frac{dy}{dx}=0$
$\Rightarrow \left(x+y\frac{dy}{dx}\right)^{^2} =0\quad$(Squaring both sides)
$\Rightarrow \quad x^{2} +2xy\frac{dy}{dx}+y^{2}\left(\frac{dy}{dx}\right)^{^2}=0$
$\Rightarrow \quad-2xy\frac{dy}{dx}=x^{2}+y^{2} \left(\frac{dy}{dx}\right)^{^2}$
$y^{2}+x^{2}\left(\frac{dy}{dx}\right)^{^2} -2xy\frac{dy}{dx}$
$=\quad x^{2}+y^{2}+x^{2}\left(\frac{dy}{dx}\right)^{^2}+y^{2}\left(\frac{dy}{dx}\right)^{^2}$
$\quad$(Adding $y^{2}+x^{2}\left(\frac{dy}{dx}\right)^{^2}$ both sides)
$\Rightarrow \quad\left(y-x\frac{dy}{dx}\right)^{^2} =a^{2}\left[1+\left(\frac{dy}{dx}\right)^{^2} \right]$
$\quad\quad\quad\quad\quad\quad\quad\quad\left(\because x^{2}+y^{2}=a^{2}\right)$