Question

For the circle ${{\left( x+3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=10,$ how do you find the equation of the tangent line at the point $\left( -2,5 \right)?$

Answer 1

Hint: We find the derivative $\dfrac{dy}{dx}$ of the given equation of circle. We calculate the value of the derivative at the given point. Then we use the general form of the equation of the tangent line.

Complete step by step solution:
Consider the given equation of circle ${{\left( x+3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=10$
We use the implicit differentiation to find the derivative of the equation of the circle.
We write,
$\Rightarrow \dfrac{d}{dx}\left( {{\left( x+3 \right)}^{2}}+{{\left( y-2 \right)}^{2}} \right)=\dfrac{d}{dx}10$
 We will get the following,
$\Rightarrow \dfrac{d}{dx}{{\left( x+3 \right)}^{2}}+\dfrac{d}{dx}{{\left( y-2 \right)}^{2}}=0$ Since the derivative of a constant is zero.
Now the next step is as follows,
$\Rightarrow 2\left( x+3 \right)\dfrac{d}{dx}\left( x+3 \right)+2\left( y-2 \right)\dfrac{d}{dx}\left( y-2 \right)=0$ Since, $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}.$
Now we differentiate inside the bracket, we get
$\Rightarrow 2\left( x+3 \right)+2\left( y-2 \right)\dfrac{dy}{dx}=0$ By linearity property of differentiation.
When we transform all the terms except $\dfrac{dy}{dx},$ we get
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{2\left( x+3 \right)}{2\left( y-2 \right)}$
We are cancelling the common factor $2$ from both the numerator and the denominator,
 $\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( x+3 \right)}{\left( y-2 \right)}$
Now, we have to find the value of the derivative of the equation at the given point $\left( -2,5 \right).$
Then we get,
$\Rightarrow {{\left[ \dfrac{dy}{dx} \right]}_{\left( -2,5 \right)}}=-\dfrac{-2+3}{5-2}=\dfrac{1}{3}.$
That is, the derivative \[\dfrac{dy}{dx}\] of the equation of circle at the given point $\left( -2,5 \right)$ is $\dfrac{1}{3}.$
The general form of the equation of the tangent line is given by, $y\left( x \right)={{y}_{0}}+{y}'\left( {{x}_{0}} \right)\left( x-{{x}_{0}} \right).$
Now we are applying the values of the derivative obtained and the coordinate values of the given point.
We take, ${{x}_{0}}=-2,$ and ${{y}_{0}}=5.$
Also, we have the derivative ${y}'\left( {{x}_{0}} \right)=\dfrac{dy}{dx}=\dfrac{1}{3}.$
Then the required equation of the tangent line can be obtained by doing the procedure given below,
$\Rightarrow y\left( x \right)=5-\dfrac{1}{3}\left( x+2 \right).$
We open the bracket to get,
$\Rightarrow y\left( x \right)=5-\dfrac{1}{3}x+\dfrac{2}{3}$
From this, we get
$\Rightarrow y\left( x \right)=-\dfrac{x}{3}+\dfrac{15-2}{3}=-\dfrac{x}{3}+\dfrac{13}{3}$
That is,
$\Rightarrow y\left( x \right)=\dfrac{13-x}{3}$

Hence, the equation of the tangent line at $\left( -2,5 \right)$ is $y\left( x \right)=\dfrac{13-x}{3}.$

Note: Given is an equation of a circle with centre $\left( -3,2 \right)$ and radius $\sqrt{10}.$
Now the line passing through $\left( -2,5 \right)$ and the centre $\left( -3,2 \right)$ is $l=\dfrac{\Delta y}{\Delta x}=\dfrac{5-2}{-2+3}=3.$
This line is perpendicular to the tangent line.
 Therefore, the tangent $m=-\dfrac{1}{3}.$
By point-slope formula, $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right).$
We get $y-5=-\dfrac{1}{3}\left( x+2 \right)=-\dfrac{x}{3}-\dfrac{2}{3}.$
Therefore, $y=-\dfrac{x}{3}+\dfrac{13}{3}.$