/*! 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
No, the equation \((x-3)^{2}+(y-5)^{2}=-25\) does not represent a circle because the radius squared of a circle cannot be a negative number in the real number system. Therefore, this equation does not present any real set in the coordinate plane.

Step by step solution

01

Identify the standard form of a circle's equation

The general formula of a circle is \( (x-a)^{2} + (y-b)^{2} = r^{2} \), where (a, b) is the center and r is the radius.
02

Compare given equation with standard form

The given equation is \((x-3)^{2}+(y-5)^{2}=-25 \). If we compare it with the standard form, this looks like a circle's equation, but the RHS (right hand side) is -25, which supposed to be \( r^{2} \), is negative. In a real number system, the square of a real number (which the radius is) can never be negative.
03

Conclusion

Since the square of a radius cannot be negative in the real number system and the provided formula has a negative value on the RHS, it's confirmed that this equation does not represent a circle. So, the given equation perhaps does not represent any geometric shape in the real coordinate plane, as it has no real solutions.

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