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Question 1126641: A circle that has its center in the fourth quadrant touches the y-axis and intersects the x-axis at (3,0) and (9,0). The area of the part of the circle in the first quadrant is...
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The circle has center (6, -3 sqrt (3))
That is required for the circle to touch the y-axis. The radius is 6 units, and the triangles made by having the center go to the x-intercepts have length 6 and the distance to the y-axis is 3 sqrt(3) by the Pythagorean theorem, given that the intersection point is (6, 0). That makes the segment of each on the y-axis 3, and makes the two central angles 30 degrees each, since 3 is half of 6, and the sine of 30 is 1/2.
the area of the sector bounded by the x-intercepts and the center is 1/6 of the area of the whole circle (36 pi). The area of the part below the y-axis has length 6 and altitude 3 sqrt(3), the distance to the center from the y-axis. That area is 3*3 sqrt (3) or 9 sqrt(3)
Therefore, the area of the segment above the y-axis is 6 pi-9 sqrt (3), which is a reasonable answer. This is 3.26 units using approximations.
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