Problem 26 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the equation of the circle in standard form. Then identify its center and radius. $$9 x^{2}+9 y^{2}+54 x-36 y+17=0$$

Step-by-Step Solution

Answer

Standard form: \((x + 3)^2 + (y - 2)^2 = 4 \). Center: (-3, 2). Radius: 2

Step 1: Divide by the coefficient of \( x^2 \) and \( y^2 \)

First divide every term in the equation by 9 to normalize the coefficients of \( x^2 \) and \( y^2 \) to 1. Doing this gives us the equation \( x^2 + y^2 + 6x - 4y + 17/9 = 0 \)

Step 2: Reorder and group terms

Rearrange and group \( x \) and \( y \) terms together. We'll group the \( x \) terms, \((x^2 + 6x)\), and the \( y \) terms, \((y^2 - 4y)\), together to get \( (x^2 + 6x) + (y^2 - 4y) = -17/9 \)

Step 3: Complete the square

Complete the square for both the \( x \) and \( y \) groups. For the \( x \) group, take half of the coefficient of \( x \), square it and add it to both sides of the equation. For the \( y \) group, do the same. This gives \((x^2 + 6x + (6/2)^2) + (y^2 - 4y + (-4/2)^2) = -17/9 + (6/2)^2 + (-4/2)^2 \). Then simplify to get \((x + 3)^2 + (y - 2)^2 = 4 \)

Step 4: Identify the center and radius

For a circle with equation \((x - h)^2 + (y - k)^2 = r^2\), the center is at (h, k) and the radius is \( r \). Therefore by comparison, for the equation \((x + 3)^2 + (y - 2)^2 = 4\), the center is at (-3, 2) and the radius is \( \sqrt{4} \), which is 2

Key Concepts

Standard Form

The standard form of a circle's equation looks like this: \((x - h)^2 + (y - k)^2 = r^2\). Here, \(h\) and \(k\) represent the center of the circle, while \(r\) indicates the radius. This form helps easily identify the key characteristics of a circle, such as its position and size.

In the given equation \(9x^2 + 9y^2 + 54x - 36y + 17 = 0\), we need to transform it to standard form. The first step is to ensure the coefficients of \(x^2\) and \(y^2\) are both 1. This is accomplished by dividing the entire equation by 9.

Upon simplifying, we get: \(x^2 + y^2 + 6x - 4y + 17/9 = 0\). Now, it's easier to see the grouping of terms, which will allow us to complete the square for both \(x\) and \(y\). This prepares us to easily rewrite the equation as a standard form circle equation.

Completing the Square

Completing the square is a technique used to simplify quadratic expressions into a perfect square form. It helps in transforming the general equation of a circle to the standard form.

Starting with the grouped equations \((x^2 + 6x) + (y^2 - 4y) = -17/9\), complete the square for both the \(x\) and \(y\) terms.

For \(x^2 + 6x\):

• Take half of 6, which is 3.
• Square it to get 9, then add and subtract 9 inside the equation.

Similarly, for \(y^2 - 4y\):

• Half of -4 is -2.
• Square it to obtain 4, then add and subtract 4 inside the equation.

Align both sides of the equation with these squares: \((x+3)^2 + (y-2)^2 = 4\). This transforms the equation into a complete square form, making it easier to identify the circle's attributes.

Center and Radius

Once the equation is in standard form, it's straightforward to identify the center and radius of the circle. In the form \((x - h)^2 + (y - k)^2 = r^2\),

• The center \((h, k)\) of the circle is found by examining the transformations applied to \(x\) and \(y\).
• The radius \(r\) is derived from the constant on the right side of the equation.

For the equation \((x + 3)^2 + (y - 2)^2 = 4\), compare it with the standard circle equation to find:

• The center is at \((-3, 2)\) due to \((x+3)\) and \((y-2)\).
• The radius is \( \sqrt{4} = 2 \).

Understanding the center and radius provides valuable information about the circle's position on the graph and the distance from the center to any point on the perimeter.