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
$2x+2y\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}=0$
$\Rightarrow {\left(x+y\frac{dy}{dx}\right)}^{{}^2}=0$(Squaring both sides)
$\Rightarrow x^2+2xy\frac{dy}{dx}+y^2{\left(\frac{dy}{dx}\right)}^{{}^2}=0$
$\Rightarrow -2xy\frac{dy}{dx}=x^2+y^2{\left(\frac{dy}{dx}\right)}^{{}^2}$
$y^2+x^2{\left(\frac{dy}{dx}\right)}^{{}^2}-2xy\frac{dy}{dx}$
$=x^2+y^2+x^2{\left(\frac{dy}{dx}\right)}^{{}^2}+y^2{\left(\frac{dy}{dx}\right)}^{{}^2}$
(Adding $y^2+x^2{\left(\frac{dy}{dx}\right)}^{{}^2}$ both sides)
$\Rightarrow {\left(y-x\frac{dy}{dx}\right)}^{{}^2}=a^2\left[1+{\left(\frac{dy}{dx}\right)}^{{}^2}\right]$
$\left(\because x^2+y^2=a^2\right)$