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Math Help - Coordinates of intersection of a line and circle?

  1. #1
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    Coordinates of intersection of a line and circle?

    How do you get it? I've tried for 30 minutes now...

    y=x-3
    x^2+y^2=65


    Yes I do know substitution, plus add, and the third way to find intersections.
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  2. #2
    o_O
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    Why don't you show us what you've done and we'll see where you went wrong?
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    Quote Originally Posted by o_O View Post
    Why don't you show us what you've done and we'll see where you went wrong?
    This is one of the attempts I made that I could find on paper.

    sqrt(65) = 8.06 = radius

    I substituted (x-3)^2 for y^2
    x^2 + (x-3)^2 = 65
    x^2 + x^2 - 6s + 9 -9 = 65 -9
    (2x^2-6x)2 = 56/2
    x^2-3x=28
    x(2x-3)=28
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    Not a bad effort. Now, you don't want to end up with your last line unless you have a 0 on the right hand side as that allows you to equate each factor equal to 0 which we don't have here.

    What you should do is try to get your expression in the form of a quadratic ( ax^{2} + bx + c = {\color{red}0}). The 0 is important as this is the key for using the quadratic formula or to factor for the reason mentioned above:
    x^{2} + x^2 - 6x + 9 = 65
    2x^{2} - 6x - 56=0
    x^{2} - 3x - 28 = 0 (divided both sides by 2)

    Can you see what to do now?
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    Quote Originally Posted by o_O View Post
    Not a bad effort. Now, you don't want to end up with your last line unless you have a 0 on the right hand side as that allows you to equate each factor equal to 0 which we don't have here.

    What you should do is try to get your expression in the form of a quadratic ( ax^{2} + bx + c = {\color{red}0}). The 0 is important as this is the key for using the quadratic formula or to factor for the reason mentioned above:
    x^{2} + x^2 - 6x + 9 = 65
    2x^{2} - 6x - 56=0
    x^{2} - 3x - 28 = 0 (divided both sides by 2)

    Can you see what to do now?
    Thanks!

    Did you get (7,4) and (-4,7), too?
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    Quote Originally Posted by AlphaRock View Post
    I substituted (x-3)^2 for y^2
    x^2 + (x-3)^2 = 65
    x^2 + x^2 - 6s + 9 -9 = 65 -9
    (2x^2-6x)2 = 56/2
    x^2-3x=28
    x(2x-3)=28
    Hello, AlphaRock! Your work looks fine so far (except for the 2 in the last step). Now let us solve it!

    Are you familiar with solving quadratic equations? There are a few methods you can use:

    First, you can try factoring:

    x^2-3x=28\Rightarrow x^2-3x-28=0

    \Rightarrow x^2+(4-7)x+(-7)\cdot4=0

    \Rightarrow (x+4)(x-7)=0

    Next, if you can't figure out how to factor the expression, or if the expression is simply too messy to easily factor, you can use the quadratic formula, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, where a, b,\text{ and }c are the coefficients of the 2nd-degree, 1st-degree, and constant terms, respectively.

    In this case, we have
    x^2-3x=28\Rightarrow x^2-3x-28=ax^2+bx+c=0

    So, in the formula, a=1, b=-3,\text{ and }c=-28.

    Alternatively, you can complete the square. For example:

    x^2-3x=28\Rightarrow x^2-3x+\frac94=28+\frac94

    \Rightarrow\left(x-\frac32\right)^2=\frac{121}4

    \Rightarrow x-\frac32=\pm\sqrt{\frac{121}4}

    \Rightarrow x=\frac32\pm\frac{11}2

    I hope that helps!

    Edit: Beaten by o_O! I'm too slow with my LaTeX!
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  7. #7
    o_O
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    Looks good.
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  8. #8
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    Ah, here's a more harder question:

    4x+y+11=0 and (x-5)^2+(y-3)^2=16


    What do I do to solve this question...? I'm clueless on what ot do next.

    Here's what I did:

    y=(-11-4x)

    (x-5)^2 + (-11-4x-3)^2 = 16
    x^2-10x+25+(-14-4x)^2 = 16
    x^2-10x+25 + 196 + 16x^2 + 112x = 16
    17x^2+102x+221-16 = 16-16
    17x^2+102x+205=0

    >>Imaginary. What do I do...? THis is all my work.
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    Quote Originally Posted by AlphaRock View Post
    4x+y+11=0 and (x-5)^2+(y-3)^2=16

    What do I do to solve this question...? I'm clueless on what ot do next.

    Here's what I did:

    y=(-11-4x)

    (x-5)^2 + (-11-4x-3)^2 = 16
    x^2-10x+25+(-14-4x)^2 = 16
    x^2-10x+25 + 196 + 16x^2 + 112x = 16
    17x^2+102x+221-16 = 16-16
    17x^2+102x+205=0

    >>Imaginary. What do I do...? THis is all my work.
    Your work is okay so far. If you continue, using the quadratic formula, you should get

    x=-\frac{2\sqrt{221}\,\text{i}+51}{17},\ \frac{2\sqrt{221}\,\text{i}-51}{17};\ <br />
y=\frac{8\sqrt{221}\,\text{i}+17}{17},\ -\frac{8\sqrt{221}\,\text{i}-17}{17}

    Edit: But, make sure you have the right problem, because I notice that if you change the addition in the second equation to a subtraction the two graphs will intersect.
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