Problem 57 - Thomas Calculus

Question

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}+4 x+y^{2}=12 $$

Step-by-Step Solution

Answer

The center is (-2, 0) and radius is 4.

Step 1: Complete the Square for x-Terms

To rewrite the given equation in the standard form of a conic section, we'll start by completing the square for the terms involving \(x\). The given equation is \(x^{2}+4x+y^{2}=12\). First, focus on the \(x\)-terms: \(x^2 + 4x\). We take half of the coefficient of \(x\), which is \(2\), square it to get \(4\), and add it inside the equation. The equation becomes \(x^2 + 4x + 4 + y^2 = 12 + 4\) or \((x+2)^2 + y^2 = 16\).

Step 2: Identify the Conic Section

With the equation transformed to \((x+2)^2 + y^2 = 16\), we recognize this as the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.Here, the center \((h, k) = (-2, 0)\) and the radius \(r = \sqrt{16} = 4\).

Step 3: Summarize Parts for a Circle

Since the equation represents a circle, there are no foci, vertices, or asymptotes typically involved. Only the center and radius are relevant for a circle.- Center: \((-2, 0)\)- Radius: \(4\)

Key Concepts

Completing the Square

Completing the square is a useful algebraic method that helps to rewrite quadratic expressions in a more manageable form. For a term like \( x^2 + 4x \), the idea is to transform it into a perfect square trinomial, which is an expression that can be factored into the square of a binomial. Here is how you can do it:

• Take the coefficient of the \( x \) term, which is \( 4 \), and divide it by \( 2 \). This gives \( 2 \).
• Square this result to get \( 4 \).
• Add and subtract this square inside the equation to complete the square. This allows you to maintain balance in the equation.

This process changes \( x^2 + 4x \) into \( (x+2)^2 - 4 \), allowing us to rewrite the original equation in a form that may reveal important characteristics of the conic section, such as its center or radius.Understanding completing the square simplifies equations, especially when dealing with conic sections like circles, parabolas, and ellipses.

Equation of a Circle

The equation of a circle in its standard form is \( (x-h)^2 + (y-k)^2 = r^2 \). Here, \( (h, k) \) represents the center of the circle while \( r \) is the radius. This form helps identify fundamental properties of a circle quickly once the equation is arranged correctly.
For example, given the equation \( (x+2)^2 + y^2 = 16 \), we can see that it is similar to the standard form by comparing terms. The \( (x+2)^2 \) suggests that the \( x \)-coordinate of the center has been shifted from \( 0 \) to \( -2 \), and \( y^2 \) remains unchanged indicating \( y = 0 \). Hence, the center of the circle is \( (-2, 0) \). Similarly, since \( 16 \) equals \( r^2 \), this value can be used to find the radius by taking the square root.This transformation into the circle equation shows the equation's simplicity and elegance, making it easy to extract and compute properties like the center and radius.

Center of a Circle

The center of a circle is a critical point that defines its position in a coordinate system. Knowing how to identify a circle's center from its equation is crucial. With the circle's equation in standard form \( (x-h)^2 + (y-k)^2 = r^2 \), \((h, k)\) directly represents the coordinates of the center.
In the exercise, the equation \((x+2)^2 + y^2 = 16\) clearly shows that the circle is centered at \((-2, 0)\). The term \((x+2)^2\) suggests a horizontal shift to the left by \(2\) units, which translates to \(-2\) for the center's \(x\)-coordinate. Meanwhile, the \(y^2\) implies no vertical shift from zero.

• The center\((h, k)\) becomes \((-2, 0)\).

The center is crucial not only for graphing the circle but also for understanding its symmetry and alignment within a plane, serving as a reference point for geometric interpretations.

Radius of a Circle

The radius of a circle is a fundamental measurement that depicts the distance from the center to any point on the circle's edge. In the equation \((x-h)^2 + (y-k)^2 = r^2\), \(r\) represents this radius.
Internally, the equation shows the radius squared as \( r^2 \). To find \( r \), simply take the square root of this value. For example, in the equation \((x+2)^2 + y^2 = 16\), compare it with the standard form. The \( 16 \) represents \( r^2 \), and thus taking its square root gives \( r = 4 \).

• From the equation, identify \( r^2 \), then solve by square root.
• For \((x+2)^2 + y^2 = 16\), \( r = \sqrt{16} \).
• The result is \( r = 4 \).

This straightforward calculation clarifies the circle's size, enabling the determination of its circumference and the area if needed.