Question 1147335
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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.<br>
Then, since the circle "touches" (is tangent to) the y-axis, the radius of the circle is 6.<br>
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.<br>
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.<br>
Use what you know about area formulas to find the answer.