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Question 486127: Find all points on the x-axis that are 5 units from the point (4,-3).
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! The locus of all the points that are located 5 units away from the point (4, -3) is a circle with the center located at the point (4, -3) and having a radius of 5. The equation for such a circle is:
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The two points where this circle intersects with the x-axis will have "y" values of zero. (Any point on the x-axis has zero as its corresponding "y" value.) And these are the two points that we are looking for.
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That being the case, set y = 0 in the circle equation and the equation reduces to:
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Square out each of the terms in this equation to get:
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Combine the two constants on the left side and this equation becomes:
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Subtract 25 from both sides:
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On the left side the +25 and -25 sum to zero, and the same thing happens on the right side. This reduces the equation to:
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Factor out an x on the left side and the equation becomes:
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Note that this equation will be true if either of the factors is equal to zero. This is because when one of the factors is equal to zero, the left side gets multiplied by zero, and this makes the left side equal to the zero on the right side.
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Therefore, the equation will be correct if either:
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or
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and in this second equation when you add + 8 to both sides it becomes:
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This means that the points 0 and +8 on the x-axis are 5 units from the point (4, -3). In the form of ordered pairs the answers will be that the points (0, 0) and (8, 0) are the two points that are on the x-axis and are 5 units from the given ordered pair (4, -3).
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Hope this helps you to understand this way of using the distance formula in the form of the equation for a circle to solve this problem.
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