Question 1203378
Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.
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Find the area of the region enclosed by the graph of the equation x^2-14x+3y+70=15+9y-y^2 that lies below the line y=x-3.
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x^2-14x+3y+70=15+9y-y^2 is a circle.  Find the center and the radius:
x^2-14x+49 + y^2-6y+9 = 15 - 70 + 58 = 3
--> (x-7)^2 + (y-3)^2 = 3
Center at (7,3):  r = sqrt(3)
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Find the distance from the center to the line x-y - 3 = 0
d = |1 - 1 -3|/sqrt(1+1) = 3/sqrt(2) = 3sqrt(2)/2
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Find the area of the minor segment, i.e., the area above the given line:
h = distance from the line to the center of the segment
h = 3 - 3sqrt(2)/2
Area = r^2*acos(1-h/r) - (r-h)*sqrt(2rh - h^2) ---- (angle in radians)
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h = ~ 3 - 2.12132 = 0.87868
Sub and find the area of the segment, then subtract that from the area of the circle.