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Find the area of the region satisfying the inequality  x^2 + y^2 \leq 4x + 6y+13. 

 Feb 8, 2019
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\(x^2 + y^2 \leq 4x + 6y+13\\ x^2-4x + y^2-6y \leq 13\\ (x^2-4x )+ (y^2-6y) \leq 13\\ (x^2-4x+4 )+ (y^2-6y+9) \leq 13+4+9\\ (x-2 )^2+ (y-3)^2 \leq (\sqrt{26})^2\\\)

 

 

 

Oh you can find the area yourself. Just use the formula.  :)

 Feb 8, 2019
edited by Melody  Feb 8, 2019

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