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+1.

Akhain1 Aug 10, 2024

#1**+1 **

we can start by moving all the terms of the equation onto the left side:

x^2-14x+55-6y+y^2=0

then we can factor:

(x^2-14x+49)+(y^2-6y+9)-3=0

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

this gives us the equation of a circle with center (7, 3) and radius √3

the circle is always below the line, so we can just find the area of the circle, which is just πr^2, which gives us 3π

doeansodc Aug 10, 2024