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Math Help - Geometry Problem (may possibly require calculus, but probably not)

  1. #1
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    Geometry Problem (may possibly require calculus, but probably not)

    Here is the problem:
    Consider the circle x^2 + y^2 - 2sqrt(3)x -2y +3 = 0 and two lines y=1/sqrt(3)x and y=sqrt(3)x drawn on the same Cartesian coordinate system. What is the area of the region that is bounded b the two lines and the circle, and that is lying in the exterior of the circle?

    I know the lines are 30 and 60 degrees from the positive x axis (trisect the first quadrant) and the circle is radius 1 with center at (sqrt(3),1), but I still can't figure this out. Please help
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  2. #2
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    Quote Originally Posted by machack View Post
    Here is the problem:
    Consider the circle x^2 + y^2 - 2sqrt(3)x -2y +3 = 0 and two lines y=1/sqrt(3)x and y=sqrt(3)x drawn on the same Cartesian coordinate system. What is the area of the region that is bounded b the two lines and the circle, and that is lying in the exterior of the circle?

    I know the lines are 30 and 60 degrees from the positive x axis (trisect the first quadrant) and the circle is radius 1 with center at (sqrt(3),1), but I still can't figure this out. Please help
    Find the points of intersection between the lines and the circle.
    Draw the diagram. Find the combination of areas under straight line, the curve and the x-axis between the limits to get the desired area.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Find the points of intersection between the lines and the circle.
    Draw the diagram. Find the combination of areas under straight line, the curve and the x-axis between the limits to get the desired area.
    Well, if i understand you correctly, you basically gave me a generic answer that stated the obvious. I wouldn't be posting here if it could be calculated so simply like that.
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  4. #4
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    Quote Originally Posted by machack View Post
    Well, if i understand you correctly, you basically gave me a generic answer that stated the obvious. I wouldn't be posting here if it could be calculated so simply like that.
    OK.
    First find out the four points of intersection of two straight lines with the circle. Do you know how to that?
    Then I will tell you next step.
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  5. #5
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    Ah i think i got it now. I think the answer is sqrt(3)/2 -pi/6. Could you verify that this is correct?
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  6. #6
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    Quote Originally Posted by machack View Post
    Ah i think i got it now. I think the answer is sqrt(3)/2 -pi/6. Could you verify that this is correct?
    Wat is that answer? Is it the required area? How iss it possible to solve so quickly? can you show your calculations?
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  7. #7
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    After chuggin in y=x/sqrt(3) and y=sqrt(3)x into the equation of the circle to find the intersection, we find that y=x/sqrt(3) intersects at (3sqrt(3)/2,3/2) and (sqrt(3)/2,1/2), of which we are only interested in the latter, and that y=sqrt(3)x intersects at one point (sqrt(3)/2,3/2), which implies that the line is tangent to the circle. Now, we find the area of the triangle with the vertices at (0,0), (sqrt(3)/2,1/2), and (sqrt(3)/2,3/2), which is easy and comes out to sqrt(3)/4. Now we see that the two points on the circle are 1 apart. Because the radius of the circle is 1 as well, the center of the circle and the two points form an equilateral triangle. Thus, the area we have to subtract from the triangle we had found the area of earlier (the curve part) is given by (area of circle)/6 - (area of equilateral triangle of side 1). After subtracting that from the triangle we get the answer as sqrt(3)/2 -pi/6.

    I think the key here was actually trying the problem out (instead of thinking this is gonna be hard to calculate and looking for other ways to tackle the problem, as I did) because the numbers come out nicely.

    But, I think there may be a way to solve this using calculus (ie using a polar function and finding the area from the angle pi/6 to pi/3). But my calculus is pretty rusty and I forget how to do that
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