SOLUTION: could anyone help me with these two questions? 1. Is the graph of a circle a graph of a function? (Yes or no) 2. Find the center and the radius of the circle represented by t

Algebra ->  Graphs -> SOLUTION: could anyone help me with these two questions? 1. Is the graph of a circle a graph of a function? (Yes or no) 2. Find the center and the radius of the circle represented by t      Log On


   



Question 145089: could anyone help me with these two questions?
1. Is the graph of a circle a graph of a function? (Yes or no)
2. Find the center and the radius of the circle represented by the equation
x2 + y2 + 10x – 4y – 7 = 0

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
1) The answer is NO! You can check this by using the "vertical line test" in which you draw a vertical line through the circle on the graph containing the circle.
If the vertical line passes through the graph (circle) only once, then the graph is a function. In the case of a circle, the vertical line would pass through the graph (circle) twice, so the equation represented by the graph is not a function.
2) Find the center and the radius of the circle represented by the equation:
x%5E2%2By%5E2%2B10x-4y-7+=+0
The goal is to transform this equation into the standard form of the equation for a circle with center at (h, k) and radius r, %28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2. First, add 7 to both sides of the equation.
x%5E2%2By%5E2%2B10x-4y+=+7 Now group the x-terms together and the y-terms together.
%28x%5E2%2B10x%29%2B%28y%5E2-4y%29+=+7 The next step is to "complete the square" in both the x-terms and the y-terms. You do this by adding the square of half the x-coefficient: (%2810%2F2%29%5E2+=+25) and likewise for the y-coefficient (%28-4%2F2%29%5E2+=+4) to both sides of the equation.
%28x%5E2%2B10x%2B25%29%2B%28y%5E2-4y%2B4%29+=+7%2B25%2B4 Now factor the x- and y-groups and simplify.
%28x%2B5%29%5E2%2B%28y-2%29%5E2+=+36 Compare this with the standard form %28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2
You can see that h = -5, k = 2, and r%5E2+=+36
So, the center is at (h,k) = (-5,2) and the radius, r+=+sqrt%2836%29 = 6.