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Chapter 10: Problem 18

Find the center and the radius of the circle with the given equation. Then draw the graph. $$y^{2}-6 y-1=8 x-x^{2}+3$$

Short Answer

Expert verified
The center is (4, 3) and the radius is 鈭�29.

Step by step solution

01

Rewrite the equation

Rewrite the given equation to group the terms involving x and y on one side: \[ y^{2} - 6y - 1 = 8x - x^{2} + 3 \]Rewrite this equation as: \[ x^{2} - 8x + y^{2} - 6y = 4 \]
02

Complete the square

To complete the square, start by completing it for the x-terms and y-terms separately: \[ x^{2} - 8x + (y^{2} - 6y) = 4 \]For the x-terms: \[ x^{2} - 8x = (x - 4)^{2} - 16 \]For the y-terms: \[ y^{2} - 6y = (y - 3)^{2} - 9 \]
03

Substitute completed square terms back into equation

Substitute the completed square terms back into the equation: \[ (x - 4)^{2} - 16 + (y - 3)^{2} - 9 = 4 \]Simplify the equation: \[ (x - 4)^{2} + (y - 3)^{2} = 29 \]
04

Identify the center and radius

The equation \((x - h)^{2} + (y - k)^{2} = r^{2}\) represents a circle with center \((h, k)\) and radius \(r\).From the simplified equation, compare and find that: Center \((h, k) = (4, 3)\)Radius \(r = \sqrt{29}\)
05

Draw the graph

Plot the center of the circle at point \((4, 3)\).Draw a circle with radius \(\sqrt{29}\) (approximately 5.39 units) around the center.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

equation of a circle
Understanding the equation of a circle is foundational in precalculus. A general form of a circle's equation in a Cartesian plane is given by \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center of the circle and \( r \) is the radius. This equation indicates that every point \( (x, y) \) on the circle is exactly \( r \) units away from the center. In our given exercise, the equation of the circle needs to be reformatted from \( y^2 - 6y - 1 = 8x - x^2 + 3 \) to this standard form. The next core concept, completing the square, helps in achieving that transformation.
completing the square
Completing the square is a process used to convert a quadratic equation into a perfect square trinomial. This technique is pivotal for transforming the given circle's equation into its standard form. Here's how it is done step by step for both x-terms and y-terms:

For the x-terms in the equation \( x^2 - 8x \):
  • We can rewrite it as \( x^2 - 8x = (x - 4)^2 - 16 \).
For the y-terms in the equation \( y^2 - 6y \):
  • This can be rewritten as \( y^2 - 6y = (y - 3)^2 - 9 \).
By adding and subtracting appropriate constants, these quadratic expression are transformed into perfect square trinomials. These transformed terms are then substituted back into the equation, converting it into the desired standard form: \( (x - 4)^2 + (y - 3)^2 = 29 \).
graphing circles
Once we have the circle's equation in the standard form, graphing the circle becomes straightforward. The final equation we obtained was \( (x - 4)^2 + (y - 3)^2 = 29 \), which tells us the center of the circle is at \( (4, 3) \) and its radius is \( \sqrt{29} \approx 5.39 \) units. To graph this circle:
  • Start by plotting the center point at \( (4, 3) \) on the Cartesian plane.
  • Using a compass, extend its leg to \sqrt{29}\ units (approximately 5.39 units).
  • Draw a circle with this radius around the center point.
It becomes visually apparent how every point on the circle is exactly \ \sqrt{29} \ units away from the center \( (4, 3) \). This precise understanding of placing every part of the equation helpfully eases the graphing process.

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