/*! 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 1: 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
No, the equation \( (x-3)^{2}+(y-5)^{2}=-25 \) does not represent a circle because the square of the radius is negative. The equation has no real solutions and hence does not represent any real, graphable set.

Step by step solution

01

Analyze the General Form

In the equation \((x-3)^{2}+(y-5)^{2}=-25\), it has the same form as the standard equation of a circle, namely \((x-a)^2+(y-b)^{2}=r^2\). However, the = -25 represents a negative radius squared.
02

Reason about the Negative Radius Squared

The square of a distance (in this case the radius) can never be negative because squaring a number always results in a positive value or zero, not negative. Therefore, = -25 is not possible if this was to represent a circle.
03

Determine what the Equation Represents

Given the oddity in the form (a negative square of the radius), the equation \((x-3)^{2}+(y-5)^{2}=-25\) does not represent a circle, or indeed any real graphable shape. Hence, \( (x-3)^{2}+(y-5)^{2}=-25 \) has no real solutions.

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