Problem 46 - A Graphical Approach to Precalculus with Limits

Question

Decide whether each equation has a circle as its graph. If it does, give the center and radius. $$x^{2}+8 x+y^{2}-6 y+16=0$$

Step-by-Step Solution

Answer

Yes, it's a circle with center (-4, 3) and radius 5.

Step 1: Group the x and y terms

First, group the equation by the terms involving x and y separately. This looks like:\[ (x^2 + 8x) + (y^2 - 6y) + 16 = 0 \]

Step 2: Move constant to the right-hand side

To complete the square, move the constant term 16 to the other side of the equation:\[ (x^2 + 8x) + (y^2 - 6y) = -16 \]

Step 3: Complete the square for x terms

Take the x terms \(x^2 + 8x\), and complete the square. Take \(\frac{8}{2} = 4\) and square it to get 16. Add and subtract 16 inside the x group:\[ (x^2 + 8x + 16 - 16) = ((x+4)^2 - 16) \]

Step 4: Complete the square for y terms

Take the y terms \(y^2 - 6y\), and complete the square. Take \(\frac{-6}{2} = -3\) and square it to get 9. Add and subtract 9 inside the y group:\[ (y^2 - 6y + 9 - 9) = ((y-3)^2 - 9) \]

Step 5: Rewrite the equation

Substitute the completed squares back into the equation:\[ ((x+4)^2 - 16) + ((y-3)^2 - 9) = -16 \]

Step 6: Simplify and isolate perfect squares

Combine the constants -16 and -9, then simplify the equation:\[ (x+4)^2 + (y-3)^2 - 25 = -16 \]\[ (x+4)^2 + (y-3)^2 = 25 \]

Step 7: Identify the center and radius

The equation \((x+4)^2 + (y-3)^2 = 25\) is in the form of a circle equation \((x-h)^2 + (y-k)^2 = r^2\). Here, the center \((h, k)\) is \((-4, 3)\) and the radius \(r\) is \(\sqrt{25} = 5\).

Key Concepts

Completing the Square

To rewrite equations in a form that is useful for identifying the properties of a circle, the method of "completing the square" is an essential tool. Essentially, completing the square transforms a quadratic expression into a perfect square trinomial. This transformation not only simplifies the equation but also reveals the center and radius of the circle.

Here's how you do it:

• Start by grouping the terms involving the same variable, as you saw with the terms in the original exercise involving \(x\) and \(y\).
• Next, move the constant term to the other side of the equation to focus solely on the quadratic terms.
• For each group, divide the coefficient of the linear term by 2 and then square it. These new expressions are the terms that "complete the square" because they turn each group into a perfect square trinomial.
• Make sure to add and subtract this new squared term within the group, so you can accurately rewrite the expression as a squared binomial, like \((x + 4)^2\) or \((y - 3)^2\).
• The finished squares, when rewritten within the equation, reveal a more useful form for further analysis.

Completing the square makes it possible to easily convert any quadratic expression into a binomial squared format, which is crucial for understanding the geometry of the equation.

Center and Radius

The concepts of center and radius are key when identifying a circle's equation in its standard form. Understanding these elements allows you to pinpoint the circle in a plane perfectly.

When a circle's equation is expressed in the format: \[(x-h)^2 + (y-k)^2 = r^2\]- \((h, k)\) represents the circle's center coordinates. These values tell you where the center of the circle is located in the Cartesian plane.- \(r\) is the radius, which is the distance from the center to any point on the circle.
In the original problem, the equation \((x+4)^2 + (y-3)^2 = 25\) can be understood by recognizing that:

• The terms \((x+4)^2\) and \((y-3)^2\) indicate a horizontal and vertical shift of the circle's center to the left by 4 and up by 3, respectively.
• This implies the circle's center is at \((-4, 3)\).
• The 25 on the right side of the equation represents the radius squared, meaning the radius \(r\) is \(\sqrt{25} = 5\).

Recognizing the center and radius from the circle's equation provides critical information on its position and size, foundational for graphing and analysis.

Graph of a Circle

Graphing a circle may seem complex at first, but with a proper understanding of the circle equation, it's merely a matter of putting knowledge into action.

To graph a circle from an equation:

• Start by identifying the center \((h, k)\) and the radius \(r\) from the equation in the form \((x-h)^2 + (y-k)^2 = r^2\).
• Plot the center point on the Cartesian plane. This is your starting point.
• From the center, measure and mark points at a distance equal to the radius in all cardinal directions (up, down, left, right).
• Draw the circle by connecting these points smoothly, ensuring all points are equidistant from the center.

For our example equation \((x+4)^2 + (y-3)^2 = 25\), the center is \((-4, 3)\) and the radius is 5. By plotting the center and marking boundary points 5 units from the center, you visualize the circle perfectly.
Graphing not only solidifies your understanding of the equation but also illustrates the spatial relationships defined by the circle's mathematical properties.