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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 $y=x-3$.

 Mar 7, 2024
 #1
avatar+128794 
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Simplify as

x^2 -14x + y^2 -6y =  - 55   complete the square on x, y

 

x^2  -14x + 49  + y^2 -6y + 9 =  -55 + 49 + 9

 

(x - 7)^2  + (y - 3)^2  =  3

 

y = sqrt ( 3 - (x-7)^2 ) + 3

 

This is a circle  centered at (7,3)  with a rdius of sqrt 3

 

This is probably best approximated with a graph

 

 

The area we want is the area of triangle ABC   plus the area of  major sector BC

 

(1/2) (sqrt 3)^2 sin (131.81°)  + pi * (sqrt 3)^2 * ( 360 - 131.81) /  360    ≈ 7.09

 

cool cool cool

 Mar 8, 2024

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