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

The graph is a circle with radius sqrt(28), so the area is 28*pi.

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

Thanks Tomsun and guest answerer,

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