Circles Everywhere
Find the area (in unit 2 ) bounded between the curve x ( x − 6 2 ) + y ( y − 8 ) + 2 7 6 = 0 and above the x -axis.
Give your answer to the nearest integer .
The answer is 162.
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2 solutions
T h i s i s a c i r c l e ( x − 1 8 ) 2 + ( y − 4 ) 2 = 8 2 . So the circle cuts a chord from x-axis that is at a distance of 8-4=4. So that chord length is = 4 ∗ 3 . Also the chord substance an angle of 120 degrees at the center, also the angle the segment that is below x-axis is 120. S e g m e n t a r e a = π ∗ 8 2 ∗ 3 6 0 1 2 0 − 4 ∗ 4 ∗ 3 . S o t h e a r e a a b o v e t h e x − a x i s = π ∗ 8 2 − { π ∗ 8 2 ∗ 3 6 0 1 2 0 − 4 ∗ 4 ∗ 3 . } = 1 6 1 . 7 5 . . . . . . . . 1 6 2 .
By expanding and completing the square, the function becomes ( x − 1 8 ) 2 + ( y − 4 ) 2 = 8 2 . This is a circle that is shifted up by 4. Finding the area can be done by subtracting area below x axis from the total area of circle. 64\pi-2\int_{0}^\sqrt{48}\sqrt{64-x^{2}}-4 dx . Finally the answer is 162