Problem 34 - 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. $$x^{2}+y^{2}+12 x+12 y+63=0$$

Step-by-Step Solution

Answer

The equation of the circle in the standard form is \((x+6)^2 + (y+6)^2 = 9\). The center of the circle is (-6, -6) and the radius is 3. To graph the circle, plot the center point at (-6, -6) and draw a circle with radius 3 centered at this point. The circle goes through points (-3, -6), (-6, -3), (-9, -6), and (-6, -9).

Step 1: Rewrite the equation in standard form

To rewrite the equation in standard form, we will complete the square for both x and y terms. The given equation is: \[x^2 + y^2 + 12x + 12y + 63 = 0\] Let's group the x and y terms: \[(x^2 + 12x) + (y^2 + 12y) = -63\] Next, we add and subtract the square of half the coefficients of the x and y terms inside the parentheses. This would mean adding and subtracting \((\frac{12}{2})^2 = 36\) for x, and adding and subtracting \((\frac{12}{2})^2 = 36\) for y. Be sure to balance the equation by adding these same terms to the right side. \[(x^2 + 12x + 36) + (y^2 + 12y + 36) = -63 + 36 + 36\] Now each parenthesis is a perfect square: \[((x+6)^2) + ((y+6)^2) = 9\] The equation in standard form is \((x+6)^2 + (y+6)^2 = 9\).

Step 2: Identify the center and radius

Now that we have the equation in standard form, we can easily find the center (h, k) and radius r. From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we can see that: - The center is (-6, -6) because \((x+6)^2 = (x-(-6))^2\) and \((y+6)^2 = (y-(-6))^2\). - The radius is 3 because \(r^2 = 9\), so \(r = \sqrt{9} = 3\).

Step 3: Graph the circle

To graph the circle, first plot the center point at (-6, -6). Next, draw a circle with radius 3 centered at this point. The circle goes through points (-3, -6), (-6, -3), (-9, -6), and (-6, -9).

Key Concepts

Completing the Square

Completing the square is a process used to convert quadratic expressions into a perfect square trinomial. This is particularly useful when dealing with circle equations as it allows us to express them in the standard form.
To complete the square in a given equation, follow these steps:

• Identify the quadratic terms. In our case: \(x^2 + 12x\) and \(y^2 + 12y\).
• Add and subtract the square of half the coefficient of the linear terms. For \(x^2 + 12x\), half of 12 is 6, so \(6^2 = 36\).
• Repeat for the \(y\) terms, adding and subtracting 36 again.

This transforms the equation, grouping terms into perfect squares: \((x+6)^2\) and \((y+6)^2\). We also adjust the constant on the right side of the equation to maintain equality.

Standard Form of a Circle

The standard form of a circle's equation is \[(x-h)^2 + (y-k)^2 = r^2\]where \((h, k)\) is the center of the circle and \(r\) is the radius.

This format is very useful because it immediately tells you the circle's center and radius once the equation is in this form.
For our example, after completing the square, we have \((x+6)^2 + (y+6)^2 = 9\). Here, the perfect squares indicate a center at \((-6, -6)\) and the equation on the right-hand side, 9, means the radius squared is 9.

Graphing Circles

Graphing a circle from its equation in standard form is straightforward. Once you have the center and radius:

• Mark the center of the circle on the graph. For our circle, this is at \((-6, -6)\).
• Use the radius to mark points around the center. The radius tells you how far each point on the circle is from the center.
• In our case, the radius is 3, so plot points 3 units away from the center in all directions, creating a circle.

Connect these points smoothly to graph the circle accurately. This circle will touch points like \((-3, -6)\), \((-6, -3)\), \((-9, -6)\), and \((-6, -9)\).

Center and Radius of a Circle

Identifying the center and radius of a circle from its standard form equation is crucial for graphing and understanding the circle.
From the equation \((x-h)^2 + (y-k)^2 = r^2\), we get the center as \((h, k)\) and the radius as \(r\). In our exercise, converting \((x+6)^2 + (y+6)^2 = 9\) indicates:

• The center is \((-6, -6)\).
• The radius is 3, since \(r^2 = 9\), therefore \(r = \sqrt{9} = 3\).

With this information, graphing and analyzing the circle becomes much simpler.