Problem 52 - A Graphical Approach to College Algebra

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}+4 x+y^{2}-8 y+32=0$$

Step-by-Step Solution

Answer

The equation does not represent a circle; it can't have a negative radius.

Step 1: Rearrange and Complete the Square for x

Start by rearranging the terms related to the variable \(x\), which are \(x^2 + 4x\). To complete the square, take the coefficient of \(x\), which is 4, halve it to get 2, and square it to get 4. Add and subtract 4 inside the equation: \(x^2 + 4x + 4 - 4\).

Step 2: Rearrange and Complete the Square for y

Next, rearrange the terms related to the variable \(y\), which are \(y^2 - 8y\). To complete the square, take the coefficient of \(y\), which is -8, halve it to get -4, and square it to get 16. Add and subtract 16 inside the equation: \(y^2 - 8y + 16 - 16\).

Step 3: Combine Completed Squares and Constants

Now, incorporate the completed squares: \((x^2 + 4x + 4) - 4\) and \((y^2 - 8y + 16) - 16\). You should also bring the constant term 32 into the equation: \((x + 2)^2 - 4 + (y - 4)^2 - 16 + 32 = 0\).

Step 4: Simplify the Equation

Combine all constants: \(-4 - 16 + 32 = 12\). Substitute back into the equation: \((x + 2)^2 + (y - 4)^2 = -12\). This step shows that the equation is a completed square form, but with \(-12\) on the right side.

Step 5: Check for Valid Circle Equation

In the standard circle equation form \((x-h)^2 + (y-k)^2 = r^2\), the right side must be a non-negative number \(r^2\) because \(r\) is the radius. Since \(-12\) is negative, this equation cannot represent a circle.

Key Concepts

Completing the Square

Completing the square is a technique used to simplify quadratic expressions. This can be very helpful in identifying the geometry described by an equation, such as a circle. To complete the square for a term like \(x^2 + 4x\), you follow these steps:
1. Take the coefficient of \(x\). Here, it's 4.
2. Halve the coefficient to get 2.
3. Square this result to get 4.
This means you'll add and subtract 4 within the equation, giving you \((x^2 + 4x + 4 - 4)\).
By doing this, \(x^2 + 4x + 4\) becomes \((x+2)^2\). You follow similar steps for \(y\), with the original terms \(y^2 - 8y\):
1. Take the coefficient of \(y\), which is -8.
2. Halve it to get -4.
3. Square it to 16.
By completing the square, you transform \(y^2 - 8y + 16\) into \((y-4)^2\). This helps us reshape the quadratic into a more manageable form.

Center and Radius of a Circle

Understanding the center and radius of a circle is crucial in analyzing circle equations. When you have an equation in the form \((x-h)^2 + (y-k)^2 = r^2\), it is in standard form. Here:

• \((h, k)\) gives you the center of the circle.
• \(r\) is the radius, which is the square root of \(r^2\).

If you complete the square correctly, the resulting equation should reveal these parameters. However, in the given exercise, the value on the right side is negative \(-12\), which means it cannot represent a valid circle. For a valid circle, \(r^2\) must be non-negative because the radius cannot be imaginary. Thus, checking whether the constant is positive after completing the square tells you if the equation can indeed represent a circle.

Standard Form of a Circle Equation

The standard form of a circle equation \((x-h)^2 + (y-k)^2 = r^2\) is essential for identifying the circle's properties. Each term helps us identify its geometrical features:

• \((x-h)^2\) and \((y-k)^2\) represent shifts in the circle's center from the origin to \((h, k)\).
• \(r^2\) is the square of the radius. It must be non-negative, as a circle cannot have a negative or imaginary radius.

If your equation simplifies to a form where the right side \(r^2\) is negative, it indicates that the expression cannot represent a real circle. Unlike in this exercise, correct application of completing the square and simplifying should yield a suitable \(r^2\) to design the circle's equation. Always ensure \(r^2 > 0\) for validity.