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

 Jan 26, 2020
 #1
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The graph is a circle with radius sqrt(28), so the area is 28*pi.

 Jan 26, 2020
 #2
avatar+283 
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Here is the answer:

    x^2 + y^2 ≤ 4x + 6y+13

  =x^2-4x + y^2-6y ≤ 13

  =(x-2)^2+(y-3)^2 ≤ 13+2^2+3^2

  =(x-2)^2+(y-3)^2 ≤ 26

  radius= √26

  the area for circle: (√26)^2 pi

  the answer is: 26 pi

 Jan 31, 2020
 #3
avatar+109468 
+1

Thanks Tomsun and guest answerer,

 

The asker may like to note this.

Some people here, including me, will not answer questions that are not presented properly.

 Jan 31, 2020

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