/*! 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 1 Determine the equation of a circ... [FREE SOLUTION] | 91影视

91影视

Determine the equation of a circle with centre at the origin and radius a) 4 units b) 3 units c) 12 units d) 2.6 units

Short Answer

Expert verified
The equations are a) \(x^2 + y^2 = 16\), b) \(x^2 + y^2 = 9\), c) \(x^2 + y^2 = 144\), d) \(x^2 + y^2 = 6.76\)

Step by step solution

01

Understand the Standard Equation of a Circle

The standard equation of a circle with center at the origin \((0,0)\) is given by the formula \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle.
02

Determine the Radius for Each Case

Identify the radius given in each part of the problem:a) 4 unitsb) 3 unitsc) 12 unitsd) 2.6 units
03

Plug in the Radius for Each Part

Substitute the radius into the standard equation for each case:a) For \(r = 4\), the equation is \(x^2 + y^2 = 4^2 = 16\).b) For \(r = 3\), the equation is \(x^2 + y^2 = 3^2 = 9\).c) For \(r = 12\), the equation is \(x^2 + y^2 = 12^2 = 144\).d) For \(r = 2.6\), the equation is \(x^2 + y^2 = 2.6^2 = 6.76\).
04

Write Down the Final Equations

Each part gives us a specific equation for the circle:a) \(x^2 + y^2 = 16\)b) \(x^2 + y^2 = 9\)c) \(x^2 + y^2 = 144\)d) \(x^2 + y^2 = 6.76\)

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.

Radius
The radius is a crucial part of any circle. It's the distance from the center of the circle to any point on its circumference. For a circle centered at the origin (0,0), the radius is denoted as \(r\). To calculate the radius if we know any point on the circle, use the distance formula. If a point (x, y) lies on the circle, then the radius \( r\) can be found using the equation \[ r = \sqrt{x^2 + y^2} \].
In practical terms, think of the radius as the 鈥渞each鈥 of the circle from its central point out to its edge.
Standard Equation
The standard equation of a circle is a mathematical expression that defines all the points (x, y) that make up the circle. For a circle centered at the origin (0,0), the standard equation is given by \[ x^2 + y^2 = r^2 \], where \[ r \] is the radius.
This equation states that for any point on the circle, the sum of the squares of its x and y coordinates will equal the square of the radius. This relationship helps in easily plotting the circle or understanding its geometry.
  • It is important because it simplifies many geometric problems involving circles.
  • It makes it easy to determine whether a point lies inside, on, or outside the circle.
Center at Origin
In the given problem, all circles are centered at the origin鈥攚hich is the point (0,0) on the coordinate plane. This makes calculations easier because the formula for the circle鈥檚 equation simplifies. Instead of adding or subtracting values to shift the center, the equation remains \[ x^2 + y^2 = r^2 \].
When the center is at the origin, the \( h \) and \( k \) (which represent the x and y coordinates of the circle's center in a more general equation) are both zero. So, the equation \[ (x - h)^2 + (y - k)^2 = r^2 \] simplifies directly into our standard equation.
Circle Equation Examples
To fully understand these concepts, let's look at the specific examples given in the exercise:
  • For a radius of 4 units: The equation will be \[ x^2 + y^2 = 4^2 = 16 \].
  • For a radius of 3 units: The equation becomes \[ x^2 + y^2 = 3^2 = 9 \].
  • With a radius of 12 units: It follows that \[ x^2 + y^2 = 12^2 = 144 \].
  • And for a radius of 2.6 units: The equation will be \[ x^2 + y^2 = 2.6^2 = 6.76 \].
These examples illustrate the simplicity and elegance of using the standard equation of a circle.

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

Convert each radian measure to degrees. Express your answers as exact values and as approximate measures, to the nearest thousandth. a) \(\frac{2 \pi}{7}\) b) \(\frac{7 \pi}{13}\) c) \(\frac{2}{3}\) d) 3.66 e) -6.14 f) -20

a) Explain, with reference to the unit circle, what the interval \(-2 \pi \leq \theta<4 \pi\) represents. b) Use your explanation to determine all values for \(\theta\) in the interval \(-2 \pi \leq \theta<4 \pi\) such that \(\mathrm{P}(\theta)=\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right).\) c) How do your answers relate to the word "coterminal"?

Earth is approximately 93 000 000 mi from the sun. It revolves around the sun, in an almost circular orbit, in about 365 days. Calculate the linear speed, in miles per hour, of Earth in its orbit. Give your answer to the nearest hundredth.

Consider the trigonometric equation \(\sin ^{2} \theta+\sin \theta-1=0\) a) Can you solve the equation by factoring? b) Use the quadratic formula to solve for \(\sin \theta\) c) Determine all solutions for \(\theta\) in the interval \(0 < \theta \leq 2 \pi .\) Give answers to the nearest hundredth of a radian, if necessary.

Draw a diagram of the unit circle. a) Mark two points, \(P(\theta)\) and \(P(\theta+\pi),\) on your diagram. Use measurements to show that these points have the same coordinates except for their signs. b) Choose a different quadrant for the original point, \(\mathrm{P}(\theta) .\) Mark it and \(\mathrm{P}(\theta+\pi)\) on your diagram. Is the result from part a) still true?

See all solutions

Recommended explanations on Math Textbooks

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.