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Find the area of the region enclosed by the graph of x^2+y^2 = 2x-6y+6+8x-2y+1.

 

 Jul 2, 2024
 #1
avatar+193 
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We can rewrite the given equation as follows:

 

\begin{align*} x^2+y^2 &= 2x-6y+6+8x-2y+1 \ (x^2+8x) + (y^2-8y) &= 7+6+1 \ (x^2+8x+16) + (y^2-8y+16) &= 30 \ (x+4)^{2} + (y-4)^{2} &= 30 = 5^2. \end{align*}

 

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​.

 Jul 2, 2024
 #2
avatar+1267 
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First, let's complete the square for both x and y to form the equation of a circle. We get

\(\left(x-5\right)^2+\left(y-\left(-4\right)\right)^2=\left(4\sqrt{3}\right)^2\)

 

This givies us a cricle with center of (5, -4) and a radius of \(4\sqrt3\)

 

Therefore, the area, which can be written as \(\pi r^2\), we have

\((4\sqrt3)^2 \pi = 16(3) \pi = 48 \pi \)

 

So 48pi is our amswer. 

 

Thanks! :)

 Jul 2, 2024

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