Question Statement

Find the equations of the tangent and normal to the circle $x^2+y^2+3x+2y+3=0$ at the point $(2,-1)$.

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Background and Explanation

To find the equations of the tangent and normal to a circle, we use implicit differentiation to determine the derivative $\frac{dy}{dx}$, which represents the slope of the tangent line. The normal line is perpendicular to the tangent at the point of contact, meaning its slope is the negative reciprocal of the tangent's slope.

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Solution

The equation of the circle is: $x^2+y^2+3x+2y+3=0$

To find the slope, we differentiate the equation with respect to $x$: $2x+2y\frac{dy}{dx}+3+2\frac{dy}{dx}=0$

Next, we isolate the $\frac{dy}{dx}$ terms to solve for the derivative: $(2y+2)\frac{dy}{dx}=-(2x+3)$ $\frac{dy}{dx}=-\frac{2x+3}{2y+2}$

Now, we evaluate the derivative at the given point $(2,-1)$ to find the slope of the tangent: $\begin{array}{rl}\frac{dy}{dx}& =-\frac{2(2)+3}{2(-1)+2}\\ & =\frac{-7}{0}=\infty \end{array}$

Equation of the Tangent

Since the slope of the tangent is undefined ($\infty$), the tangent line is vertical. A vertical line passing through a point $(x_1,y_1)$ is given by the equation $x=x_1$. Therefore, the equation of the tangent line at $(2,-1)$ is: $x=2\text{or}x-2=0$

Equation of the Normal

The normal line is perpendicular to the tangent line. Since the tangent is a vertical line, the normal must be a horizontal line. Mathematically, the slope of the normal is the negative reciprocal of the tangent's slope: $m_{normal}=\frac{-1}{\infty }=0$ A horizontal line passing through a point $(x_1,y_1)$ is given by the equation $y=y_1$. Therefore, the equation of the normal line at $(2,-1)$ is: $y=-1\text{or}y+1=0$

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Key Formulas or Methods Used

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ when $y$ is not explicitly isolated.
• Slope of Tangent ($m$): Evaluated as $\frac{dy}{dx}$ at the point $(x_1,y_1)$.
• Vertical Line Property: If $m=\infty$, the line is $x=x_1$.
• Horizontal Line Property: If $m=0$, the line is $y=y_1$.
• Perpendicular Slopes: The product of the slopes of two perpendicular lines is $-1$ ($m_1\cdot m_2=-1$).

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Summary of Steps

1. Differentiate the circle equation implicitly with respect to $x$.
2. Rearrange the resulting equation to solve for $\frac{dy}{dx}$.
3. Substitute the coordinates of the point $(2,-1)$ into the derivative.
4. Observe that the resulting slope is undefined, indicating a vertical tangent line ($x=2$).
5. Determine that the normal line must be horizontal ($y=-1$) because it is perpendicular to the vertical tangent.