Problem 58 - Precalculus Essentials

Question

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+12 x-6 y-4=0$$

Step-by-Step Solution

Answer

The center of the circle is at point (-6, 3) and the radius is 7 units. To graph, plot the center at point (-6,3) and draw a circle with a radius of 7 units.

Step 1: Rearrange the equation

First, group the \(x\) and \(y\) terms together, and move the constant to the right side of the equation. So, the equation becomes \((x^2+12x) + (y^2-6y) = 4\).

Step 2: Complete the square for x and y terms

To complete the square, consider the \(x^2+12x\) terms first. The number to add to make this a perfect square is \((12/2)^2=36\). So, add 36 to both sides and rewrite these terms as \((x+6)^2\). Do the same for the \(y^2-6y\) terms. The number to add to make this a perfect square is \((-6/2)^2=9\). So, add 9 to both sides and rewrite these terms as \((y-3)^2\). The equation now becomes \((x+6)^2 + (y-3)^2 = 4+36+9=49\).

Step 3: Identify the center and radius

Now the equation is in the standard form of the equation of a circle, \((x-h)^2 + (y-k)^2 = r^2\), where the center \((h,k)\) of the circle is given by taking the opposite of the coefficients of \(x\) and \(y\) in the complete square form, in this case \((-6, 3)\). The radius \(r\) of the circle is the square root of the constant term on the right, in this case \(\sqrt{49}=7\).

Step 4: Graph the equation

Plot the center point \((-6, 3)\) on the graph. Then draw a circle around this point with a radius of 7 units.Remember, when moving in the x direction, a movement towards the right is positive while towards the left is negative. In the y direction, a movement upwards is positive while downwards is negative. Graphing this will give a visual representation of the circle.

Key Concepts

Completing the Square

Completing the square is a technique used to simplify quadratic equations and make them easier to work with. When dealing with a circle's equation, completing the square helps in converting the general form into the standard form. For any quadratic expression like \(x^2 + bx\), you aim to add and subtract a specific number to turn it into a perfect square trinomial. This number is found using the formula \((b/2)^2\).

Let's take \(x^2 + 12x\) as an example:

• Calculate \((12/2)^2 = 36\).
• Add it to the expression to get \((x^2 + 12x + 36)\).
• Rewrite as \((x+6)^2\).

You do similar steps for the \(y^2 - 6y\) term as well. This procedure transforms the heavy algebra into neat expressions we can work with for circles.

Standard Form of a Circle

Once you complete the square for both the \(x\) and \(y\) terms, the equation of the circle can be transformed into its standard form. The standard form is crucial because it clearly displays the circle's properties. The equation \((x - h)^2 + (y - k)^2 = r^2\) is the standard form, where \((h, k)\) is the center and \(r\) is the radius of the circle.

For instance, turning \((x+6)^2 + (y-3)^2 = 49\) confirms the circle's standard form. Every term in this expression holds meaning, helping us easily identify the circle's center and radius.

Center of a Circle

The center of the circle is found directly from the standard form of a circle's equation, \((x - h)^2 + (y - k)^2 = r^2\). The values \(h\) and \(k\) represent the coordinates \((h,k)\) of the circle's center. They come by taking the opposite signs of the numbers inside the equations' perfect squares.

In our example with \((x+6)^2 + (y-3)^2 = 49\):

• The term \((x+6)^2\) corresponds to \(h = -6\).
• For \((y-3)^2\), \(k = 3\).
• Therefore, the center is \((-6, 3)\).

Determining the center allows you to graph the circle accurately, as it is the point around which the circle is drawn.

Radius of a Circle

The radius of a circle is the distance from its center to any point on its perimeter. In the standard form \((x - h)^2 + (y - k)^2 = r^2\), \(r^2\) is the number separated on the right side of the equation, and \(r\) is its square root.

In our given example, \((x+6)^2 + (y-3)^2 = 49\), we find:

• The right side of the equation is \(49\).
• The radius \(r\) is \(\sqrt{49} = 7\).

Knowing the radius lets you understand the size of the circle and assists in drawing it precisely on a graph.