SOLUTION: The diameter of a circle has endpoints P (0; 0) and Q(8;-4) : The equation of the circle is 1. (x - 8)squared + (y + 4)squared = square root 20 2.(x

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Question 1082830: The diameter of a circle has endpoints P (0; 0) and Q(8;-4) : The equation of the circle is
1. (x - 8)squared + (y + 4)squared = square root 20
2.(x - 4)squared + (y + 2) squared = 20
3. (x + 4) squared + (y - 2)squared = 2 square root 5
4.x2 + y2 = 4 square root 5
5.(x - 4)squared + (y + 2)squared = square root 80
please assist with this question I am like totally lost
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The diameter of a circle has endpoints P (0; 0) and Q(8;-4) : The equation of the circle is
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Use commas, not semi-colons.
endpoints P (0,0) and Q(8,-4)
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the center is the midpoint, call it C.
C is (4,-2). Do you know how to find that?
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The radius is the distance from C to either P or Q.
r^2 = 4^2 + 2^2 = 20
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The equation of a circle with its center at (h,k) and radius r is

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Which one matches that, if any?
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email via the TY note for help, or to check your work.

SOLUTION: The diameter of a circle has endpoints P (0; 0) and Q(8;-4) : The equation of the circle is 1. (x - 8)squared + (y + 4)squared = square root 20 2.(x - 4)squared + (y + 2) squar