Problem 80
Question
Does \((x-3)^{2}+(y-5)^{2}=0\) represent the equation of a circle? If not, describe the graph of this equation.
Step-by-Step Solution
Verified Answer
The equation \( (x-3)^{2}+(y-5)^{2}=0 \) represents a circle with a radius of zero. This circle is but a single point located at the coordinates (3,5).
1Step 1: Compare the equation with standard circle equation
The provided equation is \( (x-3)^{2}+(y-5)^{2}=0 \). This should be compared with the standard circle equation, \( (x-h)^{2} + (y-k)^{2} = r^{2} \). In the provided equation, \( h = 3 \), \( k = 5 \), and \( r^{2} = 0 \)
2Step 2: Deduce the radius of the circle
From the standard circle equation, the radius of the circle is gotten by taking the square root of \( r^{2} \). In this case, since \( r^{2} = 0 \), it means that the radius, \( r = \sqrt{0} = 0 \)
3Step 3: Describe the nature of the graph
With the radius being equal to zero, it is inferred that this is a circle with a radius of zero. Thus, this circle is actually a single point on the plane located at the coordinates of the centre which is \( (3,5) \)
Other exercises in this chapter
Problem 79
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