Problem 32 - Calculus: One and Several Variables

Question

Find an equation for the line tangent to the circle $$x^{2}+y^{2}+2 x-6 y-3=0$$ at the point (2,1)

Step-by-Step Solution

Answer

The equation of the line tangent to the given circle at the point (2, 1) is \(y = \frac{3}{2}x -2\)

Step 1: Find the center of the circle

Re-arrange the given equation of the circle in its standard form to find the center (h, k). \((x - h)^2 + (y - k)^2 = r^2\). Center of the circle is \((-1, 3)\)

Step 2: Calculate the slope of the radius line

The slope of the line from the center of the circle \((-1, 3)\) to the point of tangency \((2, 1)\) is \(\frac{1 - 3}{2 - (-1)} = -2/3\)

Step 3: Find the slope of the tangent line

The slope of the tangent line is the negative reciprocal of the slope of the radius line. So, the slope of tangent line is \( \frac{3}{2}\)

Step 4: Get the equation of the tangent line

Use point-slope form of a line \(y - y1 = m(x - x1)\) where m is the slope of the line and (x1, y1) is a point on the line. Substituting \(m = \frac{3}{2}\) and point of tangency \((2, 1)\), we get \(y - 1 = \frac{3}{2}(x - 2)\), simplifying this gives \(y = \frac{3}{2}x -2\)

Key Concepts

Circle Equation

The equation of a circle in its most recognizable form is called the **standard form**, which is \[(x - h)^2 + (y - k)^2 = r^2\] Here,

• \( (h, k) \) is the center of the circle
• \( r \) is the radius of the circle

In the given exercise, the circle is described by the equation:\[ x^2 + y^2 + 2x - 6y - 3 = 0 \]To identify the center and the radius, we need to rewrite it in standard form. This is done by completing the square, which involves rearranging and grouping the terms involving \(x\) and \(y\). For the given equation after transformations, we will find that the center is \((-1, 3)\) and the radius is the square root of a constant that the transformed equation will simplify to.This transformation is crucial in understanding the geometry of the circle and relationships such as tangency.

Point-Slope Form

The **point-slope form** of a line is a handy tool when you need to write the equation of a line when you know:

• The slope of the line \(m\)
• A point on the line \((x_1, y_1)\)

The general formula for the point-slope form is: \[ y - y_1 = m(x - x_1) \]
In the exercise provided, we calculated the slope of the tangent line to be \(\frac{3}{2}\), and the associated point we need to use is the point of tangency \((2, 1)\).
Thus, substituting these values into the point-slope form gives us:\[ y - 1 = \frac{3}{2}(x - 2) \]This equation represents the line that just touches or is tangent to the circle at the point \((2, 1)\).Using the point-slope form simplifies getting from the known slope and point to the equation describing the line.

Slope of a Line

The **slope of a line** essentially measures how steep the line is. Mathematically, it's defined as the ratio of the change in the \(y\)-values to the change in the \(x\)-values between two points on the line. This can be expressed as:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For our task, we needed to find the slope of the radius line of the circle, which runs between the center \((-1, 3)\) and the tangent point \((2, 1)\).The calculated slope turned out to be:\[ \frac{1-3}{2-(-1)} = -\frac{2}{3} \]
Understanding how to calculate the slope of a line is crucial when dealing with geometric figures like circles, as it helps determine characteristics such as tangents, where precision and understanding of geometry are necessary.

Reciprocal Slope

When two lines are perpendicular to each other, their slopes have a unique relationship: the slope of one line is the **negative reciprocal** of the other. This relationship is key when solving problems involving tangents to circles, as the tangent line to a circle at a certain point is perpendicular to the radius at that same point.
Given the slope of the radius was \(-\frac{2}{3}\), the slope of the tangent line can be found as its negative reciprocal:\[ \frac{3}{2} \]
This calculation changes the direction of the slope from descending \((-\frac{2}{3})\) to ascending \((\frac{3}{2})\), demonstrating how the line shifts its orientation upon becoming tangent.The concept of reciprocal slopes is fundamental in transformations and in visualizing how different lines interact with geometric shapes.