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After applying \(\mathbf M = \begin{pmatrix} -1 & 0 \\ -\sqrt 2 & 3 \end{pmatrix}\)to the circle of radius 3 centered at (2,0), what is the area of the resulting region?

Here's what I've done so far: I found two points - (2,0) and (-1,0), which I turned into vectors of \(\ \begin{pmatrix} 2\\ 0 \end{pmatrix}\)and \(\ \begin{pmatrix} -1\\ 0 \end{pmatrix}\) . I then multiplied these by the original matrix, to get \(\ \begin{pmatrix} -2\\ -2\sqrt2 \end{pmatrix}\) and \(\ \begin{pmatrix} 1\\ \sqrt2 \end{pmatrix}\) . I then calculated the distance between these two using the distance formula, which I got to equal \(\sqrt11\). I then plugged in \(\sqrt11\) to the circle area formula, ending up with a circle with area \(​​11\pi\). However, this answer is incorrect. Any tips on how I should solve this? Thanks!

 Mar 20, 2021
 #1
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I got it, nevermind! I calculated the distance incorrectly. The answer was \(27\pi\).

 Mar 20, 2021
 #2
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distance  between   ( -2 , -2sqrt (2) )  and   ( 1 , sqrt (2) ) =

 

sqrt  [ ( 1 - -2)^2  +  ( sqrt 2 - - 2 sqrt 2)^2  ]  = 

 

sqrt [ 3^2  + ( 3sqrt (2)) ^2 ] =

 

sqrt  [ 9 +  18 ] = sqrt (27)

 

Area =  pi  (sqrt (27) )^2  = 27 pi

 

cool cool cool

 Mar 20, 2021

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