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

#1**+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