Problem 113 - A Graphical Approach to Precalculus with Limits

Question

Describe the graph of the equation as either a circle or a parabola with horizontal axis of symmetry. Then determine two functions, designated by \(y_{1}\) and \(y_{2},\) such that their union will give the graph of the given equation. Finally, graph \(y_{1}\) and \(y_{2}\) in the same viewing rectangle. $$(x+3)^{2}+(y+1)^{2}=25$$

Step-by-Step Solution

Answer

The equation represents a circle centered at (-3, -1) with radius 5. Functions are \(y_{1} = -1 + \sqrt{25 - (x+3)^{2}}\) and \(y_{2} = -1 - \sqrt{25 - (x+3)^{2}}\).

Step 1: Identify the Equation Type

The given equation is \((x+3)^{2} + (y+1)^{2} = 25\). This is the standard form of a circle's equation \((x-h)^{2} + (y-k)^{2} = r^{2}\), where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Identify the Center and Radius of the Circle

From the standard form \((x+3)^{2} + (y+1)^{2} = 25\), we can see that the center of the circle is \((-3, -1)\) and the radius is \(\sqrt{25} = 5\).

Step 3: Rewrite the Circle Equation in Terms of Two Separate Functions

To express the circle as two functions of \(y\), solve for \(y\) in terms of \(x\):\[(y+1)^{2} = 25 - (x+3)^{2}\]\[y+1 = \pm \sqrt{25 - (x+3)^{2}}\]Thus, \(y_{1} = -1 + \sqrt{25 - (x+3)^{2}}\) and \(y_{2} = -1 - \sqrt{25 - (x+3)^{2}}\).

Step 4: Graph the Functions

Graph the functions \(y_{1} = -1 + \sqrt{25 - (x+3)^{2}}\) and \(y_{2} = -1 - \sqrt{25 - (x+3)^{2}}\) on the same set of axes. These two functions represent the upper and lower halves of the circle, respectively.

Key Concepts

Equation of a Circle

The equation given \[(x+3)^2 + (y+1)^2 = 25\]is a perfect example of the standard form of a circle's equation, which can be generally expressed as:

• \((x-h)^2 + (y-k)^2 = r^2\)

In this form, \(h\) and \(k\) represent the coordinates of the circle's center, while \(r\) represents the radius of the circle. To determine the specific elements of our equation:

• The term \((x+3)^2\) indicates a horizontal shift of -3 from the origin, giving us the x-coordinate of the center.
• Similarly, \((y+1)^2\) indicates a vertical shift of -1, providing the y-coordinate of the center.
• The number 25 on the right side of the equation represents \(r^2\), so we take its square root to find that the radius is 5.

Thus, the circle is centered at \(-3, -1\) and has a radius of 5 units.

Radius and Center

Finding the radius and center of a circle is straightforward when you recognize the format of its equation. As discussed, in the standard form \[(x-h)^2 + (y-k)^2 = r^2\]the values of \(h\) and \(k\) identify the center of the circle. Let's see how this applies:

• Our exercise gives \((x+3)^2\). By comparing, we see \(h = -3\).
• Next, from \((y+1)^2\), we derive \(k = -1\).

Thus, the center of our circle is at the point \((-3, -1)\). Identifying the radius is as important:

• The equation's right side, 25, equals \(r^2\).
• Finding \(r\) involves taking the square root: \(r = \sqrt{25} = 5\).

With a center at \(-3, -1\) and a radius of 5, you've fully characterized the circle.

Graphing Functions

Graphing a circle using its equation can initially seem complex. However, breaking it down into two functions simplifies this task. When the given equation\[(x+3)^2 + (y+1)^2 = 25\]is solved in terms of \(y\), we get:

• \((y+1)^2 = 25 - (x+3)^2\)
• Which results in \(y+1 = \pm \sqrt{25 - (x+3)^2}\).

From this expression, two distinct functions are derived:

• \(y_1 = -1 + \sqrt{25 - (x+3)^2}\)
• \(y_2 = -1 - \sqrt{25 - (x+3)^2}\)

These functions describe the upper and lower halves of the circle, respectively. Plotting these two functions on the same coordinate plane reveals the complete circle, with each function forming one half. This approach helps visualize how circles can be divided into symmetrical parts, making graphing and understanding circular functions more intuitive.