Question

Find the equations of tangent and normal to the circle ${{x}^{2}}+{{y}^{2}}-3x+4y-31=0$ at the point $\left( 2,3 \right)$.

Answer 1

Hint: You will need a point-slope formula of an equation of the line to find out the equation of the line and the normal. First, find out the slope of the tangent at the given point by differentiation with respect to $x$ and then use the relation between slopes of orthogonal lines to find the slope of normal. Finally, find the equation of normal using the point-slope formula.

Complete step-by-step solution:
We know from differential calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point.
The given equation of the circle is
\[{{x}^{2}}+{{y}^{2}}-3x+4y-31=0....(1)\]
Let us denote the point as $P\left( -2,3 \right)$ and let the slope of tangent be ${{m}_{t}}$.
We differentiate implicitly equation (1) with respect to $x$( $x$ is the independent variable) to find ${{m}_{t}}$,
\[\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{2}}+{{y}^{2}}-3x+4y-31 \right)=\dfrac{d}{dx}0 \\
 & \Rightarrow 2x+3y\dfrac{dy}{dx}-3+4\dfrac{dy}{dx}=0 \\
 & \Rightarrow \left( 3y+4 \right)\dfrac{dy}{dx}=3-2x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{3-2x}{2y+4} \\
\end{align}\]
So the slope of the tangent at the point $P\left( 2,3 \right)$ is ${{m}_{t}}={{\left. \dfrac{dy}{dx} \right|}_{\left( 2,3 \right)}}=\dfrac{3-4}{6+4}=\dfrac{-1}{10}$. We know from the point slope formula that the equation of line in a plane with point $\left( {{x}_{1}},{{y}_{1}} \right)$ and slope $m$ is given by
\[y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)...(2)\]
We put the values of ${{m}_{t}}$ , the point $P\left( 2,3 \right)$ in equation(2) and get
\[\begin{align}
  & y-\left( 3 \right)={{m}_{t}}\left( x-2 \right) \\
 & \Rightarrow y-3=\dfrac{-1}{10}\left( x-2 \right) \\
 & \Rightarrow x+10y=32 \\
\end{align}\]
We get find the equation of the tangent line to be \[x+10y=32\]. Let us denote the slope of normal as ${{m}_{n}}$. The tangent line and the normal line are always orthogonal to each other. We know that the product of slopes of two orthogonal line is $-1.$ So,
\[\begin{align}
  & {{m}_{t}}{{m}_{n}}=-1 \\
 & \Rightarrow \left( \dfrac{-1}{10} \right){{m}_{n}}=-1 \\
 & \Rightarrow {{m}_{n}}=10 \\
\end{align}\]
Now we shall use the point slope formula for equation of line to find out the equation of normal. We put the values $m={{m}_{n}}=10$ and the point $P\left( 2,3 \right)$ in equation(2) and get,
\[\begin{align}
  & y-\left( 3 \right)=10\left( x-2 \right) \\
 & \Rightarrow y-3=10\left( x-2 \right) \\
 & \Rightarrow 10x-y=17 \\
\end{align}\]
So the obtained equation of normal is $10x-y=17$.

Note: The key here is the use of the relation between the slopes of orthogonal lines which is their product is -1. We can also find the equation of normal first differentiating with respect to $y$ NOT with respect to $x$ and then the tangent line.