Question 1059944
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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:


<pre>
{{{x^2 + y^2}}} = {{{4x + 6y + 13}}}.  ===>

{{{x^2 + y^2 - 4x - 6y}}} = 13.

{{{(x^2 -4x) + (y^2 - 6y)}}} = 13.

{{{(x^2 -2*(2x) + 4) + (y^2 - 2*(3y) + 9)}}} = 13 + 4 + 9.

{{{(x-2)^2 + (y-3)^2}}} = 26.

The figure is the circle of the radius r = {{{sqrt(26)}}}.

Its area is {{{pi*r^2}}} = {{{pi*26}}} = 3.14*26.
</pre>

Solved.