Mathematics - Application of Derivative Question with Solution | TestHub

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