Problem 39 - Beginning and Intermediate Algebra

Question

Put the equation of each circle in the form \((x-h)^{2}+(y-k)^{2}=r^{2},\) identify the center and the radius, and graph. $$x^{2}+y^{2}-8 x+8 y-4=0$$

Step-by-Step Solution

Answer

The given equation can be rewritten in its standard form as \((x-4)^{2}+(y+4)^{2}=36\). The center of the circle is at point (4, -4) and its radius is 6 units. To graph the circle, plot the center at point (4, -4) on a coordinate plane and draw a circle with a radius of 6 units.

Step 1: Rewrite and rearrange the given equation

First, rewrite the given equation: \(x^{2}-8 x+y^{2}+8 y=4\) Separate x-terms, y-terms, and the constant: \((x^{2}-8 x)+(y^{2}+8 y)=4\)

Step 2: Complete the square for x-terms

In order to complete the square, find the value to be added to the x-terms: \(-8 / 2 = -4\) \((-4)^{2} = 16\) Add and subtract the value to complete the square: \((x^{2}-8 x+16)+(y^{2}+8 y)=4+16\) Rewrite x-terms as a binomial square: \((x-4)^{2}+(y^{2}+8 y)=20\)

Step 3: Complete the square for y-terms

Similar to step 2, find the value to be added to the y-terms: \(8 / 2 = 4\) \((4)^{2} = 16\) Add and subtract the value to complete the square: \((x-4)^{2}+(y^{2}+8 y+16)=20+16\) Rewrite y-terms as a binomial square: \((x-4)^{2}+(y+4)^{2}=36\)

Step 4: Identify the center and radius of the circle

Now that we have the standard form of the equation, \((x-4)^{2}+(y+4)^{2}=36\), we can identify the center and radius: Center: \((h, k) = (4, -4)\) Radius: \(r = \sqrt{36} = 6\)

Step 5: Graph the circle

To graph the circle, follow these steps: 1. Plot the center of the circle at point (4, -4) on a coordinate plane. 2. Draw a circle with a radius of 6 units centered at the plotted point. And we are done!

Key Concepts

Completing the Square

Completing the square is a method used to transform quadratic equations into a perfect square trinomial, which simplifies solving and graphing. To complete the square, follow these steps:

• Identify the quadratic term. For example, in the expression \(x^2 - 8x\), \(x^2\) is the quadratic term.
• Divide the coefficient of the linear term by 2. Here, the number is \(-8 / 2 = -4\).
• Square the result: \((-4)^2 = 16\).
• Add and subtract the squared number in the expression to complete the square, so: \(x^2 - 8x + 16\) becomes \((x - 4)^2\).

This process restructures a quadratic expression to a format that makes it easier to represent as a circle equation by converting it into a binomial square.

Standard Form of a Circle

The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where

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

This form highlights the symmetry of the circle about its center. For instance, transforming the equation \(x^2 - 8x + y^2 + 8y = 4\) using the completing the square method gives you \((x-4)^2 + (y+4)^2 = 36\). This equation immediately tells you the center is at the point \((4, -4)\), and the radius \(r\) is \(\sqrt{36} = 6\).

Understanding this form is crucial for graphing, as it provides clear visual information about the circle's location and size on the coordinate plane.

Graphing a Circle

Once you have an equation in the standard form, graphing a circle becomes straightforward. Here's how you can do it efficiently:

• Start by identifying the center \((h,k)\) and radius \(r\) from the standard form of the equation. For example, in \((x-4)^2 + (y+4)^2 = 36\), the center is \((4, -4)\) and the radius is 6.
• Plot the center point on the graph based on its coordinates.
• Using the radius, measure and mark a point from the center in all four main directions (left, right, up, and down) which will help you get the bounds of the circle.
• Draw the circle by connecting these boundary points smoothly, ensuring it is equidistant from the center at all points.

This methodical approach to graphing circles makes it easier to visualize and represents correctly the geometry of the given circle equation.