Question 1147335: A circle that has its centre in the fourth quadrant touches the y-axis and intersects the x-axis at (3,0) and (9,0). What is the area of the part of the circle in the first quadrant?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Because the x-intercepts are at 3 and 9, the center of the circle has to be on the line x=6; so the x coordinate of the center of the circle is 6.
Then, since the circle "touches" (is tangent to) the y-axis, the radius of the circle is 6.
So the distance along the x-axis between the two points of tangency is 6; and the radii of the circle to the two points of tangency are both 6. So those two radii and the portion of the x-axis they cut off form an equilateral triangle, with all angles measuring 60 degrees.
Then the area of the portion of the circle in the first quadrant is the area of a 60-degree sector of a circle with radius 6, minus the area of an equilateral triangle with side length 6.
Use what you know about area formulas to find the answer.
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