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    Q15 only, ignore Q2
    Can someone make any sense out of this?
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    You want to calculate the distance between Q and P using the pythagorean/distance formula and then half it, because the radius is half the diamater.
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    This isn't the only way to do it I don't think, but can you find the midpoint of those two points?
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    (Original post by NiamhM1801)
    This isn't the only way to do it I don't think, but can you find the midpoint of those two points?
    (1,1) , I got that but don't know what to do next.
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    (Original post by Zacken)
    You want to calculate the distance between Q and P using the pythagorean/distance formula and then half it, because the radius is half the diamater.
     a^2 + b^2 = c^2 ?
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    (Original post by ckfeister)
     a^2 + b^2 = c^2 ?
    You need to go back and learn how to calculate the distance between two points.
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    (Original post by Zacken)
    You need to go back and learn how to calculate the distance between two points.
    I just did  a^2 + b^2 = c^2 and got c = 10, radius is = 5, its the correct formula.
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    (Original post by ckfeister)
    (1,1) , I got that but don't know what to do next.
    Then find the distance from (1,1) to either one of the other 2 points - P would be easier.

    You want to find the difference in x values and square it, then the difference in y values and square that. Add these together. Take the square root, and you should get your answer
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    (Original post by NiamhM1801)
    Then find the distance from (1,1) to either one of the other 2 points - P would be easier.

    You want to find the difference in x values and square it, then the difference in y values and square that. Add these together. Take the square root, and you should get your answer
    thx
 
 
 
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