91Ó°ÊÓ

Completing the square is a technique used in algebra to convert a quadratic expression into a perfect square trinomial. This is very useful when transforming the equation of a circle into standard form. Here’s a quick summary on how to complete the square for both x and y terms in the given exercise:

• First, you group the terms involving x together and those involving y together.
• Next, identify any common coefficients among the grouped terms and factor them out.
• Then, to complete the square, you add and subtract the same value inside the parentheses. This value is \(\frac{(b/2)}{a}\textsuperscript{2}\), where b is the coefficient of the linear term and a is the coefficient of the quadratic term.
• After adding and subtracting the necessary term, you rewrite the resulting trinomial as a square of a binomial.

In our example, after grouping the terms and factoring any coefficients, we completed the square for both x and y terms, which helped rewrite the equation into a form that’s closer to the standard form of a circle.

The standard form of a circle's equation is \((x - h)\textsuperscript{2} + (y - k)\textsuperscript{2} = r\textsuperscript{2}\). This form makes it very easy to identify the center and radius of the circle. The variables h and k in the formula represent the x and y coordinates of the circle's center, while r represents the radius.
Getting our equation into this form involved completing the square as described earlier. Once the quadratic forms of x and y (i.e. \((x + \frac{2}{3})\textsuperscript{2}\) and \((y - 1)\textsuperscript{2}\)) were achieved, we identified the constants and moved them to the right side. Finally, dividing the entire equation by 9 presented a neat, standard circle equation.

• Our equation \((x + \frac{2}{3})\textsuperscript{2} + (y - 1)\textsuperscript{2} = \frac{1}{9}\) is now in standard form.
• This transformation allows us to easily see the geometric properties of the circle.

Identifying the radius and center of a circle is straightforward once the equation is in standard form \((x - h)\textsuperscript{2} + (y - k)\textsuperscript{2} = r\textsuperscript{2}\). In this form, the center of the circle is \((h, k)\) and the radius is r.
For our example, after converting the given equation to standard form \((x + \frac{2}{3})\textsuperscript{2} + (y - 1)\textsuperscript{2} = \frac{1}{9}\):

• The center of the circle is \((h, k) = (-\frac{2}{3}, 1)\).
• The radius is found by taking the square root of \(\frac{1}{9}\), giving us r = \frac{1}{3}\.

Being able to spot these values directly from the equation is key to understanding the circle’s properties. Whether you are graphing the circle or using it in applications, knowing the center and radius is essential. This is why transforming a circle's equation to standard form is such a powerful step in solving many problems in algebra and geometry.