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:
\[\begin{gathered}
  \,\,\,\,\,\dfrac{d}{{dx}}\left( {{x^2}} \right) + \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {{a^2}} \right) \\
   \Rightarrow 2x + 2y\dfrac{{dy}}{{dx}} = 0 \\
   \Rightarrow 2y\dfrac{{dy}}{{dx}} = - 2x \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{{2y}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{x}{y} \\
\end{gathered}\]
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.