Find the area of the region enclosed by the graph of x^2+y^2 = 2x-6y+6+8x-2y+1.
We can rewrite the given equation as follows:
x2+y2=2x−6y+6+8x−2y+1 (x2+8x)+(y2−8y)=7+6+1 (x2+8x+16)+(y2−8y+16)=30 (x+4)2+(y−4)2=30=52.
Thus, the equation represents a circle centered at (−4,4) with radius 5. Therefore, the area of the enclosed region is πr^2 = π(5^2) = 25π.
First, let's complete the square for both x and y to form the equation of a circle. We get
(x−5)2+(y−(−4))2=(4√3)2
This givies us a cricle with center of (5, -4) and a radius of 4√3
Therefore, the area, which can be written as πr2, we have
(4√3)2π=16(3)π=48π
So 48pi is our amswer.
Thanks! :)