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# Question8

0
101
3
+1167

What is the area of the region enclosed by the graph of the equation \(x^2-14x+3y+70=21+11y-y^2\)   that lies below the line y=x-3?

tertre  Mar 12, 2017

#3
+4716
+6

I had no idea how to do this one, but I think I understand it now!

hectictar  Mar 12, 2017
Sort:

#1
+76946
+5

We have

x^2 - 14x + 3y + 70  + y^2 - 11y - 21  = 0

x^2 - 14x + y^2 - 8y + 49   = 0

x^2 - 14x + 49 + y^2 -8y + 16  = -49 + 49 + 16

(x - 7)^2 + (y - 4)^2  = 16

This is a circle with a radius of 4  centered at (7, 4)

Look at the graph, here : https://www.desmos.com/calculator/9aqppc3dkl

The line y = x - 3    passes through (7,4)  so the line segment bounded by the circle is the diameter of that circle......so....the region of interest is just (1/2) the area of this circle  = (pi/2)*(4)^2  =

8pi units^2  ≈ 25.13 units^2

CPhill  Mar 12, 2017
#3
+4716
+6

I had no idea how to do this one, but I think I understand it now!

hectictar  Mar 12, 2017
#2
+1167
+5

Thanks so much!

tertre  Mar 12, 2017

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