/*! 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 81 Does \((x-3)^{2}+(y-5)^{2}=-25\)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 81

Does \((x-3)^{2}+(y-5)^{2}=-25\) represent the equation of a circle? What sort of set is the graph of this equation?

Short Answer

Expert verified
The graph of the equation \((x-3)^{2}+(y-5)^{2}=-25\) does not represent a circle, instead it represents an empty set. This equation has no real solutions because a square of a real number can never be negative.

Step by step solution

01

Identify the standard form of a circle

The equation of a circle in its standard form is \((x-h)^{2}+(y-k)^{2}=r^2\), in which \((h,k)\) are the coordinates of the center of the circle and \(r\) is the radius of circle. The values of \(r^2\) or the square of radius must be non-negative. It means, \(r^2 >= 0\).
02

Analyze the given equation

In the given equation \((x-3)^{2}+(y-5)^{2}=-25\), instead of being positive or zero, the value on the right side, stating the square of radius, is negative. The value \(-25\) goes against the requirement \(r^2 >= 0\) in the standard form of a circle. Therefore, this equation cannot be for a circle.
03

Identify the graph of the equation

Equations with negative values for the radius squared have no real solutions, because the square of any real number is never negative. Thus, it represents an empty set or null set. There are no points \((x,y)\) in the Cartesian coordinate system that would satisfy this equation.

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Most popular questions from this chapter

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