Question 979932: If a circle has a diameter of 10 units and passes through the coordinates (5, 0), which of the coordinates cannot also lie on the circle's boundary, if the center of the circle must have integer coordinates?
A. (-5,0)
B. (0,0)
C. (0,5)
D. (5,10)
E. (10,5)
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If the circle of interest has a diameter of 10, it has a radius of 5. In order for this circle to pass through the point (5,0), the circle of interest must have its center on a circle centered at (5,0) with a radius of 5.
The equation of the circle centered at (5,0) is
^2\ +\ y^2\ =\ 25)
which can be written:
^2})
There are 11 possible integer values for x, 0 through 10. Substituting each x value into the above equation results in 20 different values of y considering the positive and negative roots where applicable. 12 of these points have integer y values and are therefore candidates for the center of the circle of interest.
To help you narrow this down a little, I will share with you that the circle centered at (0,0) with radius 5 contains two of the points listed in the answers. The circle centered at (5,5) with radius 5 contains two others. That eliminates all but one of your answers.
John
My calculator said it, I believe it, that settles it
Answer by Edwin McCravy(20059)  (Show Source):
You can put this solution on YOUR website!
We don't need to do it like the other tutor suggests. We don't even need to
know the equation of a circle. You may not have even studied that yet,
anyway. We just plot those points and draw three circles with diameter 10
(radius 5) with centers (0,0), (5,5) and (10,5):
So we just look and see that a circle with a diameter of 10 units and
passes through the coordinates A,C,D, and E, and all three circles
have centers with integer coordinates.
So if one of those points can't lie on a circle's boundary
with integer coefficients and radius 5, it would have to be B(0,0).
We can quit here and figure that the answer can only be B(0,0).
--------------------------
However we haven't really shown that B(0,0) can't lie on a circle's boundary
with integer coefficients and radius 5. We've only gotten it by elimination.
Let's draw a circle (in red) with radius 5 that goes through B(0,0)
and also through (5,0). Let's also draw in two radii and label
the center O.
That forms an equilateral triangle, because all sides are 5 units long.
_
Since we know that the altitude of an equilateral triangle is √3/2 times
the length of a side of the equilateral_triangle, we know that the y-
coordinate of the center O is 5 times √3/2, and is irrational.
And it's the same with the only other circle we could draw with radius
5 that goes through B(0,0) and (5,0).
Edwin
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