Problem 75 - Algebra and Trigonometry

Question

73–76 ? Sketch the graph of the equation. $$ x^{2}+y^{2}+6 x-12 y+45=0 $$

Step-by-Step Solution

Answer

The graph is a point at (-3, 6).

Step 1: Identify the Type of Conic Section

The given equation is \(x^2 + y^2 + 6x - 12y + 45 = 0\). Notice that both \(x\) and \(y\) are squared, and the coefficients of \(x^2\) and \(y^2\) are the same. This suggests that the equation represents a circle.

Step 2: Complete the Square for the x-Terms

Focus on the \(x\) terms: \(x^2 + 6x\). To complete the square, take half of the coefficient of \(x\) (which is 6), square it, and add it inside the equation. Half of 6 is 3, and \(3^2 = 9\). Rewrite and adjust the equation: \[x^2 + 6x = (x + 3)^2 - 9\]

Step 3: Complete the Square for the y-Terms

Now, focus on the \(y\) terms: \(y^2 - 12y\). Similarly, take half of -12, square it and add inside the equation. Half of -12 is -6, and \((-6)^2 = 36\). Rewrite and adjust the equation:\[y^2 - 12y = (y - 6)^2 - 36\]

Step 4: Rewrite the Equation in Standard Form

Substitute the completed square expressions back into the original equation:\[(x + 3)^2 - 9 + (y - 6)^2 - 36 + 45 = 0\]Simplify to obtain:\[(x + 3)^2 + (y - 6)^2 = 0\]

Step 5: Determine the Nature and Location of the Circle

Since the equation simplifies to \((x + 3)^2 + (y - 6)^2 = 0\), it represents a circle with center \((-3, 6)\) and radius 0. This means the circle is actually a single point at \((-3, 6)\).

Step 6: Sketch the Graph

Plot the point \((-3, 6)\) on the coordinate plane. There is no actual 'circle' in the typical sense, as the radius is zero.

Key Concepts

Graphing

Graphing conic sections can sometimes be tricky, but it helps to visualize the basic shapes and their properties. Here, we are working with a circle, one of the common conic sections along with ellipses, parabolas, and hyperbolas.

When graphing, you need to start by identifying the type of conic section from its equation. For a circle, if both the \( x^2 \) and \( y^2 \) terms have the same coefficient, you are usually looking at a circle. The next step is ensuring the equation is in standard form, which makes it easier to sketch.

The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) represent the circle's center and \( r \) is the radius. After completing the square on both the x and y terms, you can identify these parameters easily. However, in this case, the equation simplifies to a radius of zero after completing the square, indicating there is a unique point rather than a typical circle to plot. This results in a point plot at \((-3, 6)\).

Completing the Square

Understanding the method of completing the square is crucial for rewriting conic section equations in standard form. This technique helps in transforming the quadratic expressions in terms of \( x \) and \( y \), helping us to decipher the properties of circles, ellipses, or parabolas.

• First, separate the terms involving \( x \) from those involving \( y \).
• For the quadratic in \( x \), start by taking the coefficient of \( x \), divide it by 2, and then square the result.

In our example, we took half of 6 to get 3, and then squared it to add 9. This gives us \((x + 3)^2\) when completed. Apply this same approach to the quadratic in \( y \).

• For \( y^2 - 12y \), half of -12 is -6, and squaring gives 36.

This transforms \( y^2 - 12y \) into \((y - 6)^2\). Through these steps, both quadratics are transformed into perfect squares, making the equation tidy and ready for analysis.

Circle Equation

The circle equation is one of the fundamental forms seen in conic sections. A proper understanding of its equation helps to easily determine key characteristics like the center and radius.

The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \). Here, \( (h, k) \) indicates the center, and \( r \) represents the radius. Interestingly, in this exercise, after completing the square, our equation simplifies to \((x + 3)^2 + (y - 6)^2 = 0\).

This surprising result reveals that \( r^2 \) equals zero, making this 'circle' effectively a point, rather than a classic circle. The center of the circle, derived as \((-3, 6)\), is the sole point of the circle as part of the conic section.

• Remember, a circle with a zero radius implies there's no tangible circle perimeter.
• Always check whether the coefficient adjustment proper leads to practical graphs.

When the work on the equation takes you to a zero-radius outcome, it's a great reminder: mathematics can showcase different, yet meaningful, outcomes for the shapes we study!