Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the circle \(x^{2}+4 x+y^{2}+3=0\) and then find equations of the two tangent lines that pass through the origin.

Step-by-Step Solution

Answer

The tangent lines are y = √(1/3)x and y = -√(1/3)x.

Step 1: Recognize the Circle Equation

We start with the given equation of the circle: x^2 + 4x + y^2 + 3 = 0. To sketch the circle, we need to rewrite it in the standard form (x-h)^2 + (y-k)^2 = r^2, where (h, k) is the center and r is the radius of the circle.

Step 2: Complete the Square

Let's complete the square for the x terms. Rewrite the equation: x^2 + 4x + y^2 = -3. For x-terms: x^2 + 4x = (x+2)^2 - 4. So, the equation becomes: (x+2)^2 - 4 + y^2 = -3. Then, (x+2)^2 + y^2 = 1. This is the standard form of a circle where the center is (-2, 0) and the radius is 1.

Step 3: Identify Circle Center and Radius

From the equation (x+2)^2 + y^2 = 1, we identify that the center of the circle is (-2, 0) and the radius is 1. With this information, we can sketch the circle on the coordinate plane.

Step 4: Determine Tangent Points from Origin

To find the tangent lines from the origin to the circle, we need to find their slope. The tangent lines will have the form y = mx, substituting this into the circle's equation we get: (x+2)^2 + (mx)^2 = 1, which resolves to finding the correct m that allows this equation to hold as a tangent.

Step 5: Solving for Slopes of Tangent Lines

Substituting y = mx into the circle's equation gives: (x+2)^2 + m^2x^2 = 1. Solve for m: x^2(1 + m^2) + 4x + 4 = 1. This simplifies to x^2(1 + m^2) + 4x + 3 = 0. This is a quadratic in x which must have a discriminant of 0 for tangency: (4)^2 - 4(1+m^2)(3) = 0. Simplify: 16 - 12(1 + m^2) = 0, 16 = 12(1 + m^2), 4 = 3(1 + m^2), 4/3 = 1 + m^2, m^2 = 1/3, so m = ±√(1/3).

Step 6: Writing Equations of the Tangent Lines

With the slope values m = √(1/3) and m = -√(1/3), we can write the two tangent lines passing through the origin as: y = √(1/3) x and y = -√(1/3) x.

Key Concepts

Tangent Lines

A tangent line to a circle is a line that touches the circle at exactly one point. This point is known as the point of tangency. Finding the equations of tangent lines often involves using the slopes of these lines, especially when the tangents need to pass through a specific point, such as the origin.
In the case of the origin, the equation of a line can be written as a simple linear equation:

• y = mx

where "m" represents the slope of the line. The task becomes finding the correct slopes that ensure the line is tangent to the circle.
To determine the slope, we substitute this form into the circle equation and set the discriminant of the resulting quadratic equation to zero, as a tangent only touches the circle at one point.
By solving these conditions, we establish the possible slopes for the tangent lines and then use these slopes to write the equations for each tangent line.

Completing the Square

Completing the square is a commonly used technique in algebra to transform a quadratic equation into a form that more easily reveals the properties of a conic section, such as a circle.
To complete the square for a term like

• x^2 + bx

we first take half of the "b" coefficient, square it, and add and subtract that square inside the expression.
For example, in the equation

• x^2 + 4x = (x + 2)^2 - 4

Here, we completed the square by adding and subtracting

• (2)^2 = 4.

This method allows the circle equation to be rewritten in standard form, making it easier to identify the circle's center and radius, key in sketching its graph.

Coordinate Geometry

Coordinate geometry allows us to use algebraic equations to describe and analyze geometric figures. It's particularly useful in the study of circles, lines, and other curves on the coordinate plane.
In the case of every circle, its equation can be expressed as

• (x - h)^2 + (y - k)^2 = r^2

where

• (h, k)

is the center of the circle and "r" is the radius. This standard form provides a straightforward way to draw the circle by identifying its crucial parameters.
For tangent lines, coordinate geometry helps us relate the slopes and points on the line and circle, providing a robust method for analysis and calculation, as seen when determining the lines passing through the origin.

Quadratic Discriminant

Understanding the quadratic discriminant is instrumental when dealing with quadratic equations, particularly to determine the nature of their roots. The discriminant is given by the expression:

• b^2 - 4ac

where "a", "b", and "c" are the coefficients in the quadratic equation

• ax^2 + bx + c = 0.

For tangent lines to a circle, only one real solution or root implies that the line touches the circle at one point. Hence, we set the discriminant to zero for the line to truly be tangent. This requirement provides the necessary condition to find suitable slope values "m" for the tangents originating from the given point.
By solving

• (4)^2 - 4(1 + m^2)(3) = 0

in this exercise, we determined the slopes practical in writing the tangent lines' equations.