/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 In Exercises \(41-46,\) find an ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 46

In Exercises \(41-46,\) find an equation for the circle with the given center \(C(h, k)\) and radius \(a\) . Then sketch the circle in the \(x y\) -plane. Include the circle's center in your sketch. Also, label the circle's \(x\) - and \(y\) -intercepts, if any, with their coordinate pairs. $$ C(3,1 / 2), \quad a=5 $$

Short Answer

Expert verified
The circle equation is \((x-3)^2 + (y-\frac{1}{2})^2 = 25\). X-intercepts are approximately \(3 \pm \frac{\sqrt{99}}{2}\), y-intercepts are \((0, \frac{9}{2}), (0, -\frac{7}{2})\).

Step by step solution

01

Understand the Circle Equation

The equation of a circle in the Cartesian plane with center \((h, k)\) and radius \(a\) is given by \((x - h)^2 + (y - k)^2 = a^2\). Our task is to use this formula with the given values of \(h\), \(k\), and \(a\).
02

Identify Given Values

We are given the center \(C(3, \frac{1}{2})\) and the radius \(a = 5\). From this, we have \(h = 3\), \(k = \frac{1}{2}\), and \(a = 5\).
03

Substitute Values into Circle Equation

Substitute \(h = 3\), \(k = \frac{1}{2}\), and \(a = 5\) into the circle equation: \((x - 3)^2 + (y - \frac{1}{2})^2 = 5^2\). This simplifies to \((x - 3)^2 + (y - \frac{1}{2})^2 = 25\).
04

Solve for Intercepts

To find the x-intercepts, set \(y = 0\) in the equation and solve for \(x\): \((x - 3)^2 + (0 - \frac{1}{2})^2 = 25\). Similarly, for the y-intercepts, set \(x = 0\) in the equation and solve for \(y\): \((0 - 3)^2 + (y - \frac{1}{2})^2 = 25\).
05

Step 4.1: Find x-Intercepts

Solve \((x - 3)^2 + \frac{1}{4} = 25\). So \((x - 3)^2 = \frac{99}{4}\) which gives \(x - 3 = \pm \sqrt{\frac{99}{4}} = \pm \frac{\sqrt{99}}{2}\). Thus, \(x = 3 \pm \frac{\sqrt{99}}{2}\).
06

Step 4.2: Find y-Intercepts

Solve \(9 + (y - \frac{1}{2})^2 = 25\). So \((y - \frac{1}{2})^2 = 16\) which gives \(y - \frac{1}{2} = \pm 4\). Thus, \(y = \frac{1}{2} \pm 4\).
07

Identify and Format Intercept Points

The x-intercepts are around \(\left(3 + \frac{\sqrt{99}}{2}, 0\right)\) and \(\left(3 - \frac{\sqrt{99}}{2}, 0\right)\). The y-intercepts are \((0, \frac{9}{2})\) and \((0, -\frac{7}{2})\).
08

Sketch the Circle

Draw the circle with center at \((3, \frac{1}{2})\) and radius 5 in the x-y plane. Mark the intercepts and the center on the graph. Your labels should include the calculated intercept points.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Center of Circle
In the equation of a circle, the center is denoted by the coordinates \(h, k\). It's crucial because it defines where the circle is positioned on the Cartesian plane. Imagine holding a pin on a paper at this point while drawing a perfect circle around it with a string tied to a pencil. The given exercise specifies the center as \(3, \frac{1}{2}\). This means our circle is centered at the point 3 units along the x-axis and \(\frac{1}{2}\) units up the y-axis. Understanding the center is key to sketching the circle correctly since it allows you to plot your circle accurately on the x-y plane.
Radius of Circle
The radius of a circle is the distance from the center to any point on the circle. In mathematical terms, it is represented as \(a\) in the standard circle equation. In the exercise, the radius is given as 5. This means every point on the circumference of the circle is 5 units away from the center \(3, \frac{1}{2}\). To visualize, if you place your center on a graph and measure 5 units in any direction, you'll eventually define the boundary of your circle. The radius helps in determining the size of the circle: larger values mean a bigger circle, smaller values mean a smaller one. With this radius, the circle equation becomes \( (x-3)^2 + (y-\frac{1}{2})^2 = 25 \).
Cartesian Plane
The Cartesian plane is the two-dimensional grid formed by the x-axis and y-axis. It鈥檚 essential for plotting mathematical functions and geometric shapes such as circles. Each point on this plane is defined by a pair of coordinates \( (x, y) \). In our case, the coordinate system helps in locating the center of the circle and plotting its shape accurately. To sketch our circle with center \(3, \frac{1}{2}\) and radius 5, we use this plane. By moving from the center point, the radius defines the circle's boundary. The grid makes it easy to visualize and draw precise graphs depicting the circle and its intercepts.
Intercepts of Circle
Intercepts are the points where the circle intersects the x-axis and y-axis. Finding these intercepts involves setting one coordinate to zero and solving for the other.
  • For x-intercepts, set \(y = 0\): The equation becomes \( (x-3)^2 + (0-\frac{1}{2})^2 = 25 \). Solving gives x-intercepts \(3 \pm \frac{\sqrt{99}}{2}\), which are approximate points of contact with the x-axis.
  • For y-intercepts, set \(x = 0\): The equation changes to \( (0-3)^2 + (y-\frac{1}{2})^2 = 25 \). Solving this reveals y-intercepts \(\left(0, \frac{9}{2}\right) \text{ and } \left(0, -\frac{7}{2}\right)\).
Identifying intercepts is important because it helps in determining how the circle fits into the overall graph, providing more specific points to draw the circle accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a curve to be symmetric about the \(x\) -axis, the point \((x, y)\) must lie on the curve if and only if the point \((x,-y)\) lies on the curve. Explain why a curve that is symmetric about the \(x\) -axis is not the graph of a function, unless the function is \(y=0\)

Exercises \(19-28\) tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. $$ y=-\sqrt{x} \quad \text { Right } 3 $$

Find the function values in Exercises \(47-50\) . $$ \cos ^{2} \frac{\pi}{8} $$

Find the domain and graph the functions in Exercises \(15-20 .\) $$ F(t)=t /|t| $$

Assume that \(f\) is an even function, \(g\) is an odd function, and both \(f\) and \(g\) are defined on the entire real line \(R .\) Which of the following (where defined) are even? odd? $$ \begin{array}{llll}{\text { a. } f g} & {\text { b. } f / g} & {\text { c. } g / f} \\ {\text { d. } f^{2}=f f} & {\text { e. } g^{2}=g g} & {\text { f. } f \circ g} \\ {\text { g. } g \circ f} & {\text { h. } f \circ f} & {\text { i. } g \circ g}\end{array} $$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.