Problem 67 - Precalculus

Question

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$100 x^{2}+100 y^{2}-100 x+400 y+409=0$$

Step-by-Step Solution

Answer

The given equation represents a circle.

Step 1: Divide by 100

The first thing to do would be to divide the entire equation by 100 to make the equation easier to handle. This yields \(x^2 - x + y^2 + 4y + 4.09 = 0\).

Step 2: Rearrange and group terms

Then, rearrange and group the equation to prepare for completing the square: \((x^2 - x) + (y^2 + 4y) + 4.09 = 0\).

Step 3: Complete the square for x and y

Now, complete the square for each term involving \(x\) and \(y\). This gives \((x - 0.5)^2 - 0.25 + (y + 2)^2 - 4 + 4.09 = 0\). Simplify to get \((x - 0.5)^2 + (y + 2)^2 = 0.16\).

Step 4: Match to conic section form

This equation is now in the standard form of a circle, \((x - h)^2 + (y - k)^2 = r^2\), where \(h\) and \(k\) are the coordinates of the center and \(r\) is the radius of the circle.

Key Concepts

Circle

A circle is a perfectly round geometric shape represented by all the points equidistant from a single point, known as the center. In a 2D coordinate plane, this is expressed mathematically through an equation in the form \[(x - h)^2 + (y - k)^2 = r^2\]where

• \(h\) and \(k\) are the coordinates of the circle's center.
• \(r\) is the radius, or distance from center to any point on the circle.

The exercise given originally had an equation that looked complex, but once simplified, it matched the standard circle form, indicating it truly represents a circle. The process involved working towards identifying \(h\), \(k\), and \(r\) to understand the circle's size and position.

Completing the Square

Completing the square is a method used to simplify quadratic equations into a perfect square trinomial. This technique is especially useful for solving problems involving circles, as it helps reveal the circle's center \((h, k)\) and radius \(r\).
To complete the square, follow these steps:

• Take the quadratic terms, such as \(x^2 - x\), and re-arrange them.
• Add and subtract the square of half the coefficient of \(x\). For \(x^2 - x\), add and subtract 0.25 to create a perfect square \((x - 0.5)^2\).
• Do the same for the \(y\) terms: \(y^2 + 4y\) becomes \((y + 2)^2 - 4\).

This technique transforms a complicated equation into a simpler one that fits a known form, like that of a circle.

Equation of a Circle

The equation of a circle in its standard form, \[(x - h)^2 + (y - k)^2 = r^2\],makes it easy to determine key characteristics of the circle:

• Center: Recognized directly from \((h, k)\), which shifts the circle along the horizontal and vertical.
• Radius: Found from \(r^2\). It's crucial to remember \(r\) is always positive, so take the square root if necessary.

The final result from our exercise was \[(x - 0.5)^2 + (y + 2)^2 = 0.16\].Here, \(h = 0.5\) and \(k = -2\) give the circle's center, and the radius \(r\) becomes 0.4 after taking the square root of 0.16. Understanding these forms enhances our ability to graph and interpret geometric shapes easily.