/*! 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 67 Does \((x-3)^{2}+(y-5)^{2}=-25\)... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 67

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 equation does not represent a circle. The graph of this equation is the empty set as the equation has no real solutions.

Step by step solution

01

Comparing with Standard Equation

Comparing our given equation with the standard circle equation, we have \((a,b) = (3,5)\), and \(r^{2} = -25\).
02

Checking for Validity of Radius

For the equation to represent a circle, the radius must be real and non-negative. Here, the square of the radius is -25. But squares of real numbers are always non-negative.
03

Concluding on Type of Set

Since the squared radius in our equation is negative, which cannot be true for any real radius, the equation cannot represent a real circle. Therefore, it has no graph in the real plane, and the set of solutions to this equation is the empty set.

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