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Chapter 2: Problem 80

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
The given equation represents a point, not a circle. The point is located at (3,5).

Step by step solution

01

Identify the form of the equation

The given equation is \((x-3)^{2}+(y-5)^{2}=0\). By comparing it with the standard form of a circle's equation which is \((x-a)^{2}+(y-b)^{2}=r^{2}\), it's clear that the equation is in the form of a circle's equation.
02

Determine the Radius

From the equation \((x-3)^{2}+(y-5)^{2}=0\), one can equate \(r^{2}\) to 0 and therefore conclude that the radius \(r\) is 0.
03

Interpret the results

A circle with a radius of 0 is simply a point. The equation therefore doesn't represent an actual circle (with a positive radius), but rather a single point.
04

Identify the point

Since the given equation has the same form of the circle's equation, we can say that the point is located at \((a,b)\), or in this case, at \((3,5)\).

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