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Find the area of the region enclosed by the graph of x^2+y^2=2x-6y+16+14x-4y+20

 Jun 6, 2024

Best Answer 

 #1
avatar+759 
+1

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! :)

 Jun 6, 2024
 #1
avatar+759 
+1
Best Answer

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! :)

NotThatSmart Jun 6, 2024

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