Find the area of the region enclosed by the graph of x^2+y^2=2x-6y+16+14x-4y+20
We want to put \( x^2+y^2=2x-6y+16+14x-4y+20\) in the form of a circle's equation.
Moving all terms to one side, we get \(x^2-16x+y^2+10y=36\)
Completing the square for both x and y, we get \((x-8)^2+(y+5)^2=125\)
\(\left(x-8\right)^2+\left(y+5\right)^2=\left(5\sqrt{5}\right)^2\)
From this, we get that \(5\sqrt5\) is the radius.
The area is \(\pi r^2 = \pi (125) = 125\pi\)
125pi is our answer
Thanks! :)
We want to put \( x^2+y^2=2x-6y+16+14x-4y+20\) in the form of a circle's equation.
Moving all terms to one side, we get \(x^2-16x+y^2+10y=36\)
Completing the square for both x and y, we get \((x-8)^2+(y+5)^2=125\)
\(\left(x-8\right)^2+\left(y+5\right)^2=\left(5\sqrt{5}\right)^2\)
From this, we get that \(5\sqrt5\) is the radius.
The area is \(\pi r^2 = \pi (125) = 125\pi\)
125pi is our answer
Thanks! :)