SOLUTION: Find the area of the region enclosed by the graph of the equation $x^2 + y^2 = 4x + 6y+13$. Thanks!

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Question 1059944: Find the area of the region enclosed by the graph of the equation $x^2 + y^2 = 4x + 6y+13$.
Thanks!

Answer by ikleyn(52830) About Me  (Show Source):
You can put this solution on YOUR website!
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Find the area of the region enclosed by the graph of the equation x^2 + y^2 = 4x + 6y + 13.
Thanks!
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The equation represents a circle.
All we need to do is to find its radius.
For it, let us reduce the equation to the standard form.
Apply the completing the square method:

x%5E2+%2B+y%5E2 = 4x+%2B+6y+%2B+13.  ===>

x%5E2+%2B+y%5E2+-+4x+-+6y = 13.

%28x%5E2+-4x%29+%2B+%28y%5E2+-+6y%29 = 13.

%28x%5E2+-2%2A%282x%29+%2B+4%29+%2B+%28y%5E2+-+2%2A%283y%29+%2B+9%29 = 13 + 4 + 9.

%28x-2%29%5E2+%2B+%28y-3%29%5E2 = 26.

The figure is the circle of the radius r = sqrt%2826%29.

Its area is pi%2Ar%5E2 = pi%2A26 = 3.14*26.

Solved.