Problem 29 - Algebra 2

Question

Find the center and radius of the circle with the given equation. Then graph the circle. $$ x^{2}+y^{2}-18 x-18 y+53=0 $$

Step-by-Step Solution

Answer

The circle's center is (9, 9) and the radius is \(\sqrt{109}\).

Step 1: Rearrange the Equation

First, rearrange the given equation by grouping the terms with the same variables together: \\[x^2 - 18x + y^2 - 18y + 53 = 0\] \Next, remember that the general form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Step 2: Complete the Square for x-terms

To complete the square for the \(x\)-terms, take the coefficient of \(x\), divide it by 2, and square it. \Coefficient of \(x\) = -18 \\((-18 / 2)^2 = 81\) \Add and subtract 81 inside the equation to complete the square: \\[x^2 - 18x + 81\] \This can be rewritten as \((x - 9)^2\).

Step 3: Complete the Square for y-terms

Similarly, complete the square for the \(y\)-terms: \Coefficient of \(y\) = -18 \\((-18 / 2)^2 = 81\) \Add and subtract 81 inside the equation to complete the square: \\[y^2 - 18y + 81\] \This can be rewritten as \((y - 9)^2\).

Step 4: Simplify the Equation and Find the Center and Radius

Substitute the completed squares back into the equation and simplify: \\[ (x - 9)^2 + (y - 9)^2 - 81 - 81 + 53 = 0\] \\[ (x - 9)^2 + (y - 9)^2 - 109 = 0\] \\[ (x - 9)^2 + (y - 9)^2 = 109\] \The equation is now in the standard form \((x-h)^2 + (y-k)^2 = r^2\). \Thus, the center \((h, k)\) of the circle is \((9, 9)\) and the radius \(r\) is \(\sqrt{109}\).

Step 5: Graph the Circle

To graph the circle, plot the center at point \((9, 9)\) on the coordinate plane. Then, use the radius \(\sqrt{109}\), which is approximately 10.44, to draw the circle. Start from the center and mark points approximately 10.44 units away in all directions to sketch the circle accurately.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique is extremely useful in converting equations into a form that is easier to work with for certain types of problems. For a quadratic expression in the form \( ax^2 + bx + c \), you start by focusing on the \( x \) terms.

• Take the coefficient of \( x \), divide it by 2, and then square the result.
• This result is then added and subtracted within the expression to maintain its balance.

This method is particularly essential when dealing with the equations of circles. For example, in the equation \( x^2 - 18x \), compute \((-18 / 2)^2 = 81\). Add and subtract 81:

• This transforms \( x^2 - 18x \) into \((x - 9)^2 - 81\).

Do this procedure separately for both the \( x \) and \( y \) terms for equations representing circles.

Standard Form of a Circle

The standard form of a circle's equation helps to identify the circle's key characteristics easily. A circle's equation in standard form is written as \((x - h)^2 + (y - k)^2 = r^2\). Here:

• \((h, k)\) represents the circle's center.
• \(r\) stands for the radius of the circle.

Converting an equation into this form allows you to discern these properties quickly and efficiently. Consider an equation like \( (x - 9)^2 + (y - 9)^2 = 109 \):

• The center is at \((9, 9)\).
• The radius is \(\sqrt{109}\). This transformation simplifies graphing and understanding the circle's properties.

Keep in mind, this form is derived by rewriting the equation from its general or expanded polynomial form, primarily using the completing the square method.

Graphing Circles

Graphing a circle starts with identifying its center and radius from its equation. Once the equation is in standard form \((x - h)^2 + (y - k)^2 = r^2\), the process becomes straightforward. Follow these easy steps:

• Plot the center \((h, k)\) on the coordinate plane. This serves as the pivot point.
• From the center, mark points that are \(r\) units away in every direction.

For accuracy, work clearly from the center outward, creating an even, smooth curve in all directions. For example, if the radius \(r\) is \(\sqrt{109}\), approximately 10.44, ensure all points on the circle are 10.44 units from the center \((9, 9)\). Drawing tools or software can help maintain precision for a visually accurate graph.

Center and Radius of a Circle

Understanding the center and radius of a circle is crucial for graphing and analyzing its properties. Given a circle's equation in standard form \((x - h)^2 + (y - k)^2 = r^2\):

• The center point \((h, k)\) serves as the circle's anchor. It dictates where the circle is located on the graph.
• The radius \(r\) measures the distance from the center to any point on the circle's edge. It defines the circle's size.

For example, from the equation \( (x - 9)^2 + (y - 9)^2 = 109 \), the center is \((9, 9)\), making it the focal point for plotting on a coordinate plane.Moreover, the radius \(\sqrt{109}\) (approximately 10.44) determines how far the circle extends outward from the center. Being able to determine the center and radius easily allows for precise and accurate graphing of circles.