Problem 79 - College Algebra

Question

Exercises \(77-79\) will help you prepare for the material covered in the first section of the next chapter. Complete the square and write the circle's equation in standard form: $$x^{2}+y^{2}-2 x+4 y-4$$ Then give the center and radius of the circle and graph the equation.

Step-by-Step Solution

Answer

The center of the circle is at point (1, -2) and the radius is 1.

Step 1: Re-write the equation

First, rearrange the equation by grouping the x terms and y terms together and take constants on right hand side: \(x^{2}-2x+y^{2}+4y = 4\)

Step 2: Complete the square within each of the groupings

Complete the square within the x terms and the y terms by adding and subtracting the square of half of each coefficient of x and y. Here, for \(x^{2}-2x\), take half of -2, square it and add and subtract that value from LHS. Likewise for \(y^{2}+4y\), take half of 4, square it and add and subtract from LHS. \[x^{2}-2x+1-1+y^{2}+4y+4-4 = 4.\] Then regroup and simplify the equation for the next step. \(x^{2}-2x+1+y^{2}+4y+4-1-4 = 4\) becomes \((x-1)^2 +(y+2)^2 = 1\)

Step 3: Extract the Center and Radius

Now, the equation is in the standard form of a circle's equation. From the standard form, we can now see the center (h, k) and the radius r. Here, h=1, k=-2, and r=1, so the center is (1, -2) and the radius is 1.

Step 4: Graph the Equation

To graph the equation, plot the center (1,-2) on a 2-D plane. Then, draw a circle with the radius 1 around that point. Include the points where the circle crosses the x and y-axes.

Key Concepts

Completing the Square

When given a quadratic equation, completing the square helps us rewrite it in a form that is easier to work with, particularly for geometric interpretations like circles.
It's about transforming an equation into a perfect square trinomial, which simplifies reading key features.

• Start by grouping the x and y terms separately.
• Add and subtract the square of half the coefficient of x and y terms within the group.

Let's break this down using part of the original equation: - For the term: \(x^2 - 2x\), find half of -2, which is -1. Square it to get 1. Add and subtract this value: \(x^2 - 2x + 1 - 1\) becomes \((x-1)^2\).- Similarly, for \(y^2 + 4y\), half of 4 is 2. Square it to get 4. Add and subtract: \(y^2 + 4y + 4 - 4\) simplifies to \((y+2)^2\).By completing the square, we've simplified our equation, making it much easier to see the circle's key features.

Standard Form of a Circle

The standard form of a circle's equation is a simple and elegant way to represent a circle on a coordinate plane.
It has the form \((x-h)^2 + (y-k)^2 = r^2\) where:

• \(h\) is the x-coordinate of the circle's center.
• \(k\) is the y-coordinate of the circle's center.
• \(r\) represents the radius.

Once you rewrite the circle's equation into this form,
you can easily identify where the circle is on the coordinate plane and how wide it spreads.
In the exercise, we completed the square to turn the given equation into this neat format: \((x-1)^2 + (y+2)^2 = 1\). This standard form lets us see at a glance all the important circle information.

Radius and Center of a Circle

Knowing how to identify the center and radius from the standard form is key.
Look at the equation \((x-h)^2 + (y-k)^2 = r^2\).

Identifying Center:

The center is simply the point \((h, k)\).
In our example, \((x-1)^2 + (y+2)^2 = 1\), the center \(h\) and \(k\) can be directly pulled from the equation.

• Here, \(h = 1\) and \(k = -2\) making the center \((1, -2)\).

Finding the Radius:

The radius is the square root of the right side of the equation \(r^2\).

• With \(r^2 = 1\), the radius \(r\) is \(\sqrt{1} = 1\).

These values make it a breeze to sketch and understand the circle's position and size on the plane.