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Math Help - intersection of circles

  1. #1
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    intersection of circles

    (x-3)^2 + (y-1)^2 = 49
    (x+2)^2 + (y+4)^2 = 169

    I expanded the two equations, etc etc, and I got 3 intersections, but the answer is 2 intersections (I even graphed the two circles and it's 2 intersections). I tried at least 3 times, looked over what I did 6 times, but I can't find out what I did wrong.. why am I getting 3 answers when there's 2 (2 of the 3 are right, the last one is wrong).
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  2. #2
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    can you post your solution?
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    Quote Originally Posted by SpringFan25 View Post
    can you post your solution?
    (3,8) ; (10, 1) ; (-4, 1)

    (-4, 1) is not an intersection.
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  4. #4
    Super Member Quacky's Avatar
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    You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.

    Let x=4
    Then x^2=16.
    But if you then square root both sides:
    x=\pm~4

    But x\neq -4 because it contradicts the original statement!

    Do you see how an extra solution has arisen? The method in itself has no flaw, it's just that square rooting is a "one-to-many" operation. To verify solutions, you have to substitute into the original equations to check that they do indeed work. Please post your method and we can check for errors.
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    Quote Originally Posted by Quacky View Post
    You may be introducing an extraneous solution. This occurs frequently with squaring/rooting.

    Let x=4
    Then x^2=16.
    But if you then square root both sides:
    x=\pm~4

    But x\neq -4 because it contradicts the original statement!

    Do you see how an extra solution has arisen? The method in itself has no flaw, it's just that square rooting is a "one-to-many" operation. To verify solutions, you have to substitute into the original equations to check that they do indeed work. Please post your method and we can check for errors.
    I substituted and I still got the right answer (and there were no square roots used). Here's my full solution.

    (x-3)^2 + (y-1)^2 = 49 EQ. 1
    (x+2)^2 + (y+4)^2 = 169 EQ. 2

    Expanded EQ. 1 and 2 to get EQ. 3 and EQ. 4 respectively.

    Subtracted EQ 4 from EQ 3 and got x = 11-y

    Subsituted x = 11-y into EQ. 3 and got y = 8 and y = 1.

    Substitued y = 8 into EQ. 3 and got x = 3.

    Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.

    I hope that's not too hard to follow.
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  6. #6
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    Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.
    This is where the issue is. x=-4 is not a solution to your original system of equations. Its might be a solution, but it doesn't have to be.

    You want the intersection of EQ 3. and the line y=11-x

    Its true that an intersection occurs at y=1, but because EQ3 is a circle there are two possible x values at that y value. You need to check which one(s) you're interested in by seeing if they satisfy the original equations.

    Graph on wolfram alpha
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  7. #7
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    Quote Originally Posted by SpringFan25 View Post
    This is where the issue is. x=-4 is not a solution to your original system of equations. Its might be a solution, but it doesn't have to be.

    You want the intersection of EQ 3. and the line y=11-x

    Its true that an intersection occurs at y=1, but because EQ3 is a circle there are two possible x values at that y value. You need to check which one(s) you're interested in by seeing if they satisfy the original equations.

    Graph on wolfram alpha
    Wait, what? I had x=11-y. How would I know when to change it to y=11-x?
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  8. #8
    Super Member Quacky's Avatar
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    Quote Originally Posted by iragequit View Post
    I substituted and I still got the right answer (and there were no square roots used). Here's my full solution.

    (x-3)^2 + (y-1)^2 = 49 EQ. 1
    (x+2)^2 + (y+4)^2 = 169 EQ. 2

    Expanded EQ. 1 and 2 to get EQ. 3 and EQ. 4 respectively.

    Subtracted EQ 4 from EQ 3 and got x = 11-y

    Subsituted x = 11-y into EQ. 3 and got y = 8 and y = 1.

    Substitued y = 8 into EQ. 3 and got x = 3.

    Substitued y = 1 into EQ. 3 and got x = 10 and x = -4.

    I hope that's not too hard to follow.
    It would have been nicer if you'd replied with a full method so I wouldn't have had to have done every single calculation for myself.

    3) x^2+y^2-6x+-2y=39
    4) x^2+4x+y^2+8y=149

    4)-3)
    10x+10y=110
    x+y=11
    y=11-x

    Sub. into:

    1) (x-3)^2 + (y-1)^2 = 49
    <br />
(x-3)^2 + (10-x)^2 = 49

    x^2-6x+9+x^2-20x+100=49

    2x^2-26x+60=0

    x^2-13x+30=0
    <br />
(x-3)(x-10)=0

    x=3 or x=10
    <br />
y=11-x

    y=8 or y=1

    The extraneous solution is in this stage:
    'Substitued y = 1 into EQ. 3 and got x = 10 and x = -4'

    Because equation 3 is a polynomial involving squared terms, so there are possibly two solutions for any given x value that will satisfy equation 3, but that doesn't mean they'll satisfy equation 4. Notice how in my method I deliberately avoided this by using the y=11-x instead of equation 3 for the last step? This was deliberately to avoid the extraneous solutions. Do you see why your method gave the extra solution?

    x^2+y^2-6x-2y=39
    If you let x=3,
    3^2+y^2-6(3)-2y=39
    9+y^2-18-2y=39
    y^2-2y-48=0
    (y-8)(y+2)=0
    y=8 or y=-2
    ...are generated. Both solutions work for equation 3, but they don't necessarily work for equation 4. Why would they? You have only substituted into equation 3 after all!

    So you can either check both in equation 4, or use the previously derived linear function to only obtain the valid solutions.
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  9. #9
    Super Member Quacky's Avatar
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    Quote Originally Posted by iragequit View Post
    Wait, what? I had x=11-y. How would I know when to change it to y=11-x?
    You don't have to! Either will work.
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  10. #10
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    Quote Originally Posted by Quacky View Post
    It would have been nicer if you'd replied with a full method so I wouldn't have had to have done every single calculation for myself.

    3) x^2+y^2-6x+-2y=39
    4) x^2+4x+y^2+8y=149

    4)-3)
    10x+10y=110
    x+y=11
    y=11-x

    Sub. into:

    1) (x-3)^2 + (y-1)^2 = 49
    <br />
(x-3)^2 + (10-x)^2 = 49

    x^2-6x+9+x^2-20x+100=49

    2x^2-26x+60=0

    x^2-13x+30=0
    <br />
(x-3)(x-10)=0

    x=3 or x=10
    <br />
y=11-x

    y=8 or y=1

    The extraneous solution is in this stage:
    'Substitued y = 1 into EQ. 3 and got x = 10 and x = -4'

    Because equation 3 is a polynomial involving squared terms, so there are possibly two solutions for any given x value that will satisfy equation 3, but that doesn't mean they'll satisfy equation 4. Notice how in my method I deliberately avoided this by using the y=11-x instead of equation 3 for the last step? This was deliberately to avoid the extraneous solutions. Do you see why your method gave the extra solution?

    x^2+y^2-6x-2y=39
    If you let x=3,
    3^2+y^2-6(3)-2y=39
    9+y^2-18-2y=39
    y^2-2y-48=0
    (y-8)(y+2)=0
    y=8 or y=-2
    ...are generated. Both solutions work for equation 3, but they don't necessarily work for equation 4. Why would they? You have only substituted into equation 3 after all!

    So you can either check both in equation 4, or use the previously derived linear function to only obtain the valid solutions.
    Wow.......... I could have just used the simple x=11-y........ WOW, how did I not think of that. So I lost a mark and wasted 5 minutes on a test doing absolutely non-sense. At least I learned, thanks guys.
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