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Chapter 13: Problem 42

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
No, \((x-3)^{2}+(y-5)^{2}=0\) doesn't represent the equation of a circle. The graph of this equation simply represents a single point located at coordinates (3,5).

Step by step solution

01

Formulate the standard equation of circle

The standard form of a circle's equation is \((x-h)^{2} + (y-k)^{2} = r^{2}\), where (h, k) is the center and r is the radius of the circle. Here, comparing the given equation \((x-3)^{2}+(y-5)^{2}=0\), it is found that h = 3, k = 5, and r \( \sqrt{0} = 0 \).
02

Determine the nature of the circle

We know that the radius of a circle can never be zero. A circle with zero radius does not hold any area. However, a circle with zero radius is a single point located at the center of the circle.
03

Describe the graph of the equation

Accordingly, the graph of the equation \((x-3)^{2}+(y-5)^{2}=0\) would not depict a circle in the usual sense but simply a single point located at coordinates (3,5).

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Most popular questions from this chapter

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$x=(y-4)^{2}-1$$

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Will help you prepare for the material covered in the next section. Solve by the addition method: $$\left\\{\begin{array}{l}2 x+4 y=-4 \\\3 x+5 y=-3\end{array}\right.$$

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