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Chapter 2: Problem 80

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
The equation \((x-3)^{2}+(y-5)^{2}=0\) represents a circle with its center at \((3, 5)\) and radius \(0\), which means it's actually a single point located at \((3, 5)\).

Step by step solution

01

Identify the coordinates of the center

Compare the given equation \((x-3)^{2}+(y-5)^{2}=0\) with the standard equation. The center of the circle can be obtained as \((h,k) = (3,5)\), where \(h\) and \(k\) represent the x and y coordinates of the center.
02

Calculate the radius of the circle

Now observe the right side of the given equation, which should be \(r^{2}\). Since this is 0, the radius \(r = \sqrt{0} = 0\).
03

Describe the graph of the equation

With these parameters, the circle has a center at \((3, 5)\) and a radius of \(0\). This means that the circle is just a point located at \((3, 5)\) in the plane.

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Most popular questions from this chapter

Begin by graphing the standard cubic function, \(f(x)-x^{3} .\) Then use transformations of this graph to graph the given function. $$ r(x)-(x-3)^{3}+2 $$

If you are given a function's graph, how do you determine if the function is even, odd, or neither?

Begin by graphing the standard cubic function, \(f(x)-x^{3} .\) Then use transformations of this graph to graph the given function. $$ r(x)-(x-2)^{3}+1 $$

What is a piecewise function?

Begin by graphing the standard cubic function, \(f(x)-x^{3} .\) Then use transformations of this graph to graph the given function. $$ h(x)-\frac 12 x^{3} $$
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