/*! 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 8 Find the center and radius of th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 8

Find the center and radius of the circle whose equation is given. $$15 x^{2}+15 y^{2}=10$$

Short Answer

Expert verified
Answer: The center of the circle is (0, 0) and the radius is \(\sqrt{\frac{2}{3}}\).

Step by step solution

01

Rewrite the equation in the standard form

To rewrite the equation in the standard form, we first need to divide both sides of the equation by 15. The standard form is \((x-a)^{2}+(y-b)^{2}=r^2\), where the center of the circle is (a, b) and r is the radius. Dividing both sides by 15 gives: $$x^{2}+y^{2}=\frac{2}{3}$$ Now, the equation is in the standard form.
02

Identify the center and radius

Since our equation is \(x^{2}+y^{2}=\frac{2}{3}\), we can see that there are no specific x or y terms. Therefore, the circle's center (a, b) is (0, 0). The radius, r, can be found by taking the square root of the constant term on the right side of the equation. In this case, r = \(\sqrt{\frac{2}{3}}\). So, the center of the circle is (0, 0) and the radius is \(\sqrt{\frac{2}{3}}\).

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!

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

The given curve is part of the graph of an equation in \(x\) and \(y .\) Find the equation by eliminating the parameter. \(x=e^{t}, \quad y=t, \quad\) any real number \(t\)

Find the equation of the ellipse that satisfies the given conditions. Center (7,-4)\(;\) foci on the line \(x=7 ;\) major axis of length \(12 ;\) minor axis of length 5.

(a) Graph these hyperbolas (on the same screen if possible \()\) $$\frac{y^{2}}{4}-\frac{x^{2}}{1}=1 \quad \frac{y^{2}}{4}-\frac{x^{2}}{12}=1 \quad \frac{y^{2}}{4}-\frac{x^{2}}{96}=1$$ (b) Compute the eccentricity of each hyperbola in part (a). (c) On the basis of parts (a) and (b), how is the shape of a hyperbola related to its eccentricity?

Locate all local maxima and minima (other than endpoints of the curve. $$x=4 t-6, \quad y=3 t^{2}+2, \quad-10 \leq t \leq 10$$

If \(a>0\) and \(b>0,\) then the eccentricity of the hyperbola $$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}+b^{2}}}{a} .\) Find the eccentricity of the hyperbola whose equation is given. $$6(y-2)^{2}=18+3(x+2)^{2}$$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

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.