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Chapter 2: Problem 60

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-x+2 y+1=0$$

Short Answer

Expert verified
The standard form of the equation is \( (x-0.5)^{2}+ (y+1)^{2}=1.25\). The center of the circle is at (0.5, -1) and the radius is approximately 1.12.

Step by step solution

01

Rearrange and Group

Group the terms that contain the same variable together. Rearrange the equation to form complete squares. The equation becomes : \( (x^{2}-x)+(y^{2}+2y)=-1 \).
02

Complete the Square - x terms

For the x terms, add the square of half the coefficient of x, which is 0.25 or (1/2)^2, to both sides. You get : \( (x^{2}-x+0.25)+ (y^{2}+2y)=0.25.\)
03

Complete the Square - y terms

Do the same for the y terms. In this case, add the square of half the coefficient of y, which is 1 or (2/2)^2, to both sides. Now the equation becomes: \( (x^{2}-x+0.25)+ (y^{2}+2y+1)=1.25.\)
04

Convert to Standard Form

Rewrite the perfect square trinomials as (variable - value)^2. So, \( (x-0.5)^{2}+ (y+1)^{2}=1.25.\)
05

Identify Center and Radius

The center of the circle is at (h, k) and the radius is \(r\), given the equation of a circle in standard form as \( (x-h)^{2}+ (y- k)^{2}= r^{2}\). Thus, the center is (0.5, -1) and \( r = \sqrt{1.25}\), which is approximately 1.12.
06

Graphing

To graph the circle, plot the center point at (0.5, -1). Draw a circle around the center point with a radius of approximately 1.12 units. Be sure to label the center point and indicate the radius on your graph.

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