Find The Area Of The Region Satisfying The Inequality $x^{2} + y^{2} \leq 4 x + 6 y + 13$ $x^2+y^2<=4x+6y+13$ Help, Plz?

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Find The Area Of The Region Satisfying The Inequality $x^{2} + y^{2} \leq 4 x + 6 y + 13$ $x^2+y^2<=4x+6y+13$ Help, Plz?

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Answer 1 · 2018-02-23T12:03:15

$13 π \approx 40.8 {units}^{2}$ $13pi~~40.8"units"^2$

Explanation:

The function can be rearranged to get:
$f (x, y) \leq 13$ $f(x,y)<=13$

Now, $f (x, y) \leq 13$ $f(x,y)<=13$ is just a form of the equation of a circle: $x^{2} - a x + y^{2} - b y = r^{2}$ $x^2-ax+y^2-by=r^2$

We will ignore what $f (x, y)$ $f(x,y)$ is since that just determines where the centre of the circle is. However, $r$ $r$ is the radius of the circle. $r = \sqrt{13}$ $r=sqrt(13)$

$Area of a circle = π r^{2}$ $"Area of a circle"=pir^2$

$r^{2} = 13$ $r^2=13$

$Area = 13 π$ $"Area"=13pi$

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Find The Area Of The Region Satisfying The Inequality $x^{2} + y^{2} \leq 4 x + 6 y + 13$ $x^2+y^2<=4x+6y+13$ Help, Plz?

1 Answer

Explanation:

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Find The Area Of The Region Satisfying The Inequality $x^{2} + y^{2} \leq 4 x + 6 y + 13$ $x^2+y^2<=4x+6y+13$ Help, Plz?