Problem 45 - A Graphical Approach to College Algebra

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}+6 x+y^{2}+8 y+9=0$$

Step-by-Step Solution

Answer

Yes, it graphs a circle with center (-3, -4) and radius 4.

Step 1: Identify the general form of a circle equation

A circle's equation is generally written as \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Let's first rearrange the given equation to try to match this form.

Step 2: Complete the square for x and y terms

Rearrange terms: \[x^2 + 6x + y^2 + 8y + 9 = 0\] Complete the square for the \(x\) terms: \[x^2 + 6x = (x+3)^2 - 9\]Complete the square for the \(y\) terms: \[y^2 + 8y = (y+4)^2 - 16\]

Step 3: Substitute completed squares back into the equation

Substitute the completed squares from Step 2 back into the equation: \[(x+3)^2 - 9 + (y+4)^2 - 16 + 9 = 0\]Simplify: \[(x+3)^2 + (y+4)^2 - 16 = 0\]

Step 4: Rewrite the equation in standard form of a circle

Add 16 to both sides to isolate the terms squared:\[(x+3)^2 + (y+4)^2 = 16\]This is the standard form of a circle with center and radius in correct format.

Step 5: Determine the center and radius

From the standard form equation \((x+3)^2 + (y+4)^2 = 16\), identify the center and the radius:- Center \((h, k)\) is \((-3, -4)\).- Radius \(r\) is \(\sqrt{16} = 4\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial, making it easier to solve and analyze. This technique is especially useful in deriving the standard form of a circle from a given equation. In essence, you want to reconfigure expressions like \(x^2 + bx\) into a form \((x+d)^2\), where \(d\) is a constant number.

To complete the square for the equation \(x^2 + 6x + y^2 + 8y + 9 = 0\), follow these steps:

• Focus on the \(x\) terms first: \(x^2 + 6x\). To complete the square, take half of the coefficient of \(x\), square it, and add/subtract it inside the equation: \(\frac{6}{2} = 3\), \(3^2 = 9\). Hence, \(x^2 + 6x\) becomes \((x+3)^2 - 9\).
• Next, for the \(y\) terms: \(y^2 + 8y\), follow the same process. \(\frac{8}{2} = 4\), \(4^2 = 16\). Thus, \(y^2 + 8y\) becomes \((y+4)^2 - 16\).

This systematically transforms the equation closer to the circle's standard form, paving the way to determine its center and radius.

Center of Circle

The center of a circle in the standard form equation \((x-h)^2 + (y-k)^2 = r^2\) is represented by the coordinates \((h, k)\). It is the point from which all points on the circle are equidistant, meaning they have the same distance from this central point.
Once we have completed the square, the equation \((x+3)^2 + (y+4)^2 = 16\) highlights the shifted "center" of the circle relative to the origin. Specifically:

• The \(x\)-coordinate of the center is found by taking the opposite of the value inside the \(x\) square term, \(-3\) in this case.
• Similarly, the \(y\)-coordinate is determined by the opposite of the \(y\) term's interior value, resulting in \(-4\).

Therefore, the center of the circle is \((-3, -4)\), a critical concept as it establishes the fixed point around which the circle is drawn.

Radius of Circle

The radius of a circle is a fixed distance from the center to any point on the circle. Finding the radius is crucial for understanding the circle's size. In the equation \((x-h)^2 + (y-k)^2 = r^2\), the symbol \(r\) represents the radius. It is derived simply by taking the square root of the constant on the other side of the equation.

In our equation \((x+3)^2 + (y+4)^2 = 16\), the constant \(16\) is equal to \(r^2\). Therefore:

• To find the radius \(r\), compute \(\sqrt{16}\).
• This results in \(r = 4\).

This means every point on the circle is 4 units away from the center point \((-3, -4)\). Understanding the radius is a key aspect of fully grasping a circle's spatial characteristics, guiding both calculation and visualization.