Question

The gradient of the tangent line at the point $\left( {a\cos \alpha ,a\sin \alpha } \right)$ to the circle ${x^2} + {y^2} = {a^2}$ is
A $\tan \left( {\pi - \alpha } \right)$
B $\tan \alpha$
C $\cot \alpha$
D $- \cot \alpha$

Answer 1

Hint:In this problem, first we need to find the derivative of the equation of the given circle with respect to $x$. Next, substitute the given points into the equation of derivative to obtain the gradient of the tangent line at a given point.

Complete step-by-step answer:
The derivative of the equation of the circle with respect to $x$ is calculated as follows:

\,\,\,\,\,\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {{a^2}} \right) \\\ \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} = 0 \\\ \Rightarrow 2y\dfrac{{dy}}{{dx}} = - 2x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{{2y}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{y} \\\ \end{gathered}$$ Now, substitute $$a\cos \alpha $$ for $$x$$ and $$a\sin \alpha $$ for $$y$$ in the above equation to obtain the gradient of the tangent line. $$\begin{gathered} \,\,\,\,\,\,\dfrac{{dy}}{{dx}} = - \dfrac{{a\cos \alpha }}{{a\sin \alpha }} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\cos \alpha }}{{\sin \alpha }} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - \cot \alpha \\\ \end{gathered}$$ Thus, the gradient of the tangent line at the point $$\left( {a\cos \alpha ,a\sin \alpha } \right)$$ is $$- \cot \alpha$$, hence, option (D) is the correct answer. Note: The first derivative of the equation of the circle with respect to $$x$$ represents the slope of the tangent line at any point on the circle.