/*! 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 80 Does \((x-3)^{2}+(y-5)^{2}=0\) r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Does \((x-3)^{2}+(y-5)^{2}=0\) represent the equation of a circle? If not, describe the graph of this equation.

Short Answer

Expert verified
No, \((x-3)^{2}+(y-5)^{2}=0\) does not represent the equation of a circle with a nonzero radius. The equation represents a point located at (3,5) in the coordinate plane.

Step by step solution

01

Identify the form of the equation

The given equation \((x-3)^{2}+(y-5)^{2}=0\) is of the form \((x-h)^{2}+(y-k)^{2}=r^{2}\). Here, \(h=3\), \(k=5\), and \(r^{2}=0\). This is the standard form of the equation of a circle where \((h, k)\) is the center and \(r\) is the radius.
02

Determine the values of h, k, and r

From the standard form, \(h=3\), \(k=5\) and \(r^{2}=0\). Taking the square root of \(r^{2}\), we find that \(r=0\).
03

Describe the graph

Since the radius of the circle is 0, the circle degenerates to a single point. Therefore, the graph of the equation will be a single point at the location \((h,k)\) or in this case, \((3,5)\).

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 91Ó°ÊÓ!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.