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
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