/*! 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 58 Find the center and the radius o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 58

Find the center and the radius of each circle. $$ x^{2}+(y-4)^{2}=11 $$

Short Answer

Expert verified
The center of the circle is at the point \((0,4)\) and the radius is \(\sqrt{11}\).

Step by step solution

01

Identify the Center

For the equation of a circle \((x-h)^{2}+(y-k)^{2}=r^{2}\), the center is given by the coordinates \((h, k)\). In the exercise equation \(x^{2}+(y-4)^{2}=11\), we have \(h=0\) and \(k=4\). Therefore, the center of the circle is at the point \((0,4)\).
02

Find the Radius

In the equation of a circle, the radius is given by \(r\), which is the square root of the value on the right side of the equation. So, for the exercise equation \(x^{2}+(y-4)^{2}=11\), the radius \(r\) is \(\sqrt{11}\).

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 a Circle
To determine the center of a circle when you are given its equation, there's a helpful formula. The standard form of the equation of a circle is given by:
  • \((x-h)^{2} + (y-k)^{2} = r^{2}\)
Here, the center of the circle is located at the point
  • \((h, k)\)
In simpler terms, the values of \(h\) and \(k\) are taken from the parts of the equation where \((x - h)\) and \((y - k)\) appear. These are your coordinates for the center of the circle.
In the exercise equation \(x^{2}+(y-4)^{2}=11\), notice there is no \((x-h)\) term, so \(h\) is 0. The term \((y-4)\) reveals that \(k = 4\). Thus, the center is
  • \((0, 4)\).
This method lets you find where the center of the circle lies swiftly.
Radius
The radius of a circle is the distance from its center out to any point on its boundary. Using the circle equation in standard form, we calculate this as the square root of the number on the right side of the equals sign.
  • Equation: \((x-h)^{2}+(y-k)^{2} = r^{2}\)
  • Radius: \(r\) is the square root of the constant on the right.
For example, in the equation \(x^{2}+(y-4)^{2}=11\), identify \(11\) as the value you need to take the square root of.
So, calculate:
  • \(r = \sqrt{11}\)
This gives us the radius. Remember, radius is always a positive value as it represents physical distance.
Coordinate Geometry
Coordinate Geometry, also known as analytic geometry, combines algebra and geometry using a coordinate system. This method allows us to solve geometry problems algebraically and understand geometric figures with the aid of coordinates.
  • In the realm of circles, equations can represent a circle on a coordinate plane.
  • Every point on a circle is equidistant from its center, a basis of this geometry.
By relating algebraic equations to coordinate geometry, you can explore circles by their equations as explored in this exercise. In our specific problem with the equation \(x^{2}+(y-4)^{2}=11\), the circle's representation involves breaking down the squared terms to find particular features like the center and radius.
Coordinate geometry empowers us to visualize, analyze, and interpret geometric shapes with precision.

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

Which equation represents a parabola that opens downward? \(F x=-2 v^{2}\) G. \(x=2 y^{2}\) H. \(y=-2 x^{2}\) \ J. \(y=2 x^{2}\)

Write an exponential equation \(y=a b^{x}\) whose graph passes through the given points. \(\left(0, \frac{1}{3}\right)\) and \((2,3)\)

Writing The area of a circle is \(\pi r^{2}\) . The area of an ellipse is \(\pi a b\) . Explain the connection.

Expand each binomial. $$ (x-y)^{3} $$

Find an equation of an ellipse for each given height and width. Assume that the center of the ellipse is \((0,0) .\) $$ h=15 \mathrm{ft}, w=32 \mathrm{ft} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

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.