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Chapter 1: Problem 48

Graph the circles whose equations are given in Exercises 47鈥�52. Label each circle鈥檚 center and intercepts (if any) with their coordinate pairs. $$ x^{2}+y^{2}-8 x+4 y+16=0 $$

Short Answer

Expert verified
The circle has a center at (4, -2) and a radius of 2, with an x-intercept at (4, 0). It has no y-intercepts.

Step by step solution

01

Identify the Circle Equation Form

The given equation is \(x^2 + y^2 - 8x + 4y + 16 = 0\). This is a general form that can be converted into standard form, \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center, and \(r\) is the radius of the circle.
02

Complete the Square for x and y

Rewrite the equation by grouping the \(x\) terms and \(y\) terms: \((x^2 - 8x) + (y^2 + 4y) = -16\). To complete the square for \(x\), add and subtract \(16\) inside the equation, and for \(y\), add and subtract \(4\): \((x^2 - 8x + 16 - 16) + (y^2 + 4y + 4 - 4) = -16\).
03

Simplify and Convert to Standard Form

The completed square form becomes \((x - 4)^2 - 16 + (y + 2)^2 - 4 = -16\). Simplifying gives \((x - 4)^2 + (y + 2)^2 = 4\), which is the standard form of a circle equation.
04

Determine the Circle's Center and Radius

From \((x - 4)^2 + (y + 2)^2 = 4\), the center of the circle is \((h, k) = (4, -2)\), and \(r^2 = 4\) implies \(r = 2\).
05

Find the Intercepts

To find the x-intercepts, set \(y = 0\): \((x - 4)^2 + (0 + 2)^2 = 4\), giving \((x - 4)^2 = 0\), so \(x = 4\). The x-intercept is \((4, 0)\). To find the y-intercepts, set \(x = 0\): \((0 - 4)^2 + (y + 2)^2 = 4\), giving \(y = -2 - \sqrt{4 - 16}\) which is non-real, so there are no y-intercepts.
06

Summary of Results

The circle has a center at \((4, -2)\) with a radius of 2. The only intercept is \((4, 0)\).

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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 essential in geometry. The equation serves as the mathematical representation of a circle in a coordinate plane. In its most recognizable form, it is expressed as
  • \[(x-h)^2 + (y-k)^2 = r^2\]
where
  • \((h, k)\) is the center of the circle, and
  • \(r\) is the radius.
This equation directly illustrates how every point \((x, y)\) on the circle is equidistant from the center. To understand how this form arises, we often start from an expanded equation which requires reorganization and manipulation to reveal the circle's center and radius.
Completing the Square
Completing the square is a crucial technique for transforming a quadratic expression into a perfect square trinomial. In the context of circle equations, it helps us reshape the general form of a circle into the standard form. In the exercise, the given equation
  • \(x^2 + y^2 - 8x + 4y + 16 = 0\)
is rewritten to group the \(x\) and \(y\) terms together, and then completed separately. The process involves:
  • Identifying the terms for each variable group, e.g., \(x^2 - 8x\).
  • Adding and subtracting the square of half of the coefficient of \(x\) and \(y\), thus \((\frac{-8}{2})^2 = 16\) for \(x\) and \((\frac{4}{2})^2 = 4\) for \(y\).
  • These adjustments yield perfect square trinomials: \((x-4)^2\) and \((y+2)^2\).
These steps convert the messy general form into a neat circle equation, making it easier to interpret.
Circle Intercepts
Intercepts are the points where a circle crosses the x-axis or y-axis. To identify these, we set one variable to zero and solve for the other. For x-intercepts, set
  • \(y = 0\)
and solve for \(x\). This results in
  • \(x = 4\)
thus the x-intercept is \((4, 0)\). Y-intercepts require setting
  • \(x = 0\)
and solving for \(y\). In the provided exercise, the attempt does not yield real solutions, indicating no y-intercepts. It is crucial to perform these checks to fully understand how the circle interacts with the coordinate axes.
Standard Form of a Circle
The standard form of a circle equation is key because it makes understanding and graphing the circle simpler. From the process of completing the square, we derived the standard form \
  • \((x-4)^2 + (y+2)^2 = 4\),
where the center is at \((4, -2)\) and the radius \(r\) is calculated from \(r^2 = 4\), giving \(r = 2\).This form offers several insights at a glance:
  • The center \((h, k)\) tells us to move 4 units in the x-direction and -2 units in the y-direction from the origin.
  • The radius \(r\) helps in determining the size of the circle.
Having the circle in standard form aids not just in graphing but also in further analyses, such as finding intercepts or confirming symmetry.

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Most popular questions from this chapter

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