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

Center: (-2, 0); Radius: 4.

Step 1: Rearrange the Equation

First, we need to rewrite the given equation in a form that is easier to understand. The equation is \(x^2 + 4x + y^2 = 12\). We'll aim to complete the square to express it in standard form for a circle.

Step 2: Complete the Square

To complete the square for the \(x\) terms, take half of the coefficient of \(x\), which is \(4\), divide it by 2 to get \(2\), then square it to get \(4\). Add and subtract this value in the equation. Thus, the equation becomes \(x^2 + 4x + 4 - 4 + y^2 = 12\).

Step 3: Simplify the Equation

Rewriting the equation incorporating the completed square, we get \((x+2)^2 - 4 + y^2 = 12\). Simplify it further by combining constants: \((x+2)^2 + y^2 = 16\).

Step 4: Identify the Conic Section Type

The simplified equation \((x+2)^2 + y^2 = 16\) is in the form of a circle. A circle is represented by the equation \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center, and \(r\) is the radius.

Step 5: Determine the Circle's Center and Radius

From \((x+2)^2 + y^2 = 16\), we identify the center as \((-2, 0)\) and the radius \(r\) as \(\sqrt{16} = 4\).

Key Concepts

Circle

A circle is one of the basic shapes in geometry. It can be defined as the set of all points that are equidistant from a fixed point, known as the center. This constant distance from the center to any point of the circle is called the radius. Circles are important because they have symmetrical properties and can be used to solve various problems in mathematics and engineering.

Every circle has several key parts:

• Center: the fixed point around which the circle is drawn.
• Radius: the distance from the center to any point on the circle.
• Diameter: twice the radius, the distance across the circle through the center.

It’s important to feel comfortable working with these properties because they are essential when solving problems related to circles in different conic sections.

Standard Form

The standard form of a circle's equation is a way of expressing the circle's properties in a clear mathematical language. It’s written as \( (x-h)^2 + (y-k)^2 = r^2 \), where

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

This equation is beneficial as it allows us to easily read off the center and radius of the circle, making it straightforward to graph or analyze. When the equation of a circle differs from this form, mathematical techniques such as completing the square can be used to transform it into standard form.

Understanding how to recognize and convert non-standard equations into this form is crucial when dealing with circles. This is especially useful when dealing with complex scenarios in conic sections.

Complete the Square

Completing the square is a valuable algebraic technique used to convert quadratic expressions into a perfect square trinomial. This method is particularly beneficial when trying to rewrite equations in their standard form, such as with circles.

Here’s how you complete the square for a quadratic expression:

• Start with a quadratic expression like \( x^2 + bx \).
• Take half of the linear coefficient \( b \), square it, and add and subtract this square inside the equation.
• Transform the quadratic part into a binomial square.

For example, to complete the square for the expression \( x^2 + 4x \), take half of 4, which is 2, and square it to get 4. This turns \( x^2 + 4x \) into \((x+2)^2\) when rearranged.By mastering this technique, you simplify the process of expressing conic sections in recognizable forms, making it easier to identify their properties.

Center and Radius

The center and radius of a circle are two of its most defining features. They are crucial for graphing the circle and solving related problems.

The center, denoted by \((h, k)\), is the fixed point from which every point on the circle is equidistant. The radius \(r\) is this constant distance.To find these from the standard form equation \((x-h)^2 + (y-k)^2 = r^2\):

• Identify \(h\) and \(k\) from the portions within the brackets \((x-h)\) and \((y-k)\).
• The radius \(r\) is the square root of the constant term on the right side of the equation.

In the exercise above, from the equation \((x+2)^2 + y^2 = 16\), you know:

• the center is \((-2, 0)\).
• the radius is \(4\) (since \(r = \sqrt{16} = 4\)).

Understanding these concepts lets you quickly visualize and understand any circle defined within a coordinate plane.