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![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 ~~ \begin{array}{lllll} center~(&\quad {{ h}},&\quad {{ k}}\quad )\qquad radius=&{{ r}}\\ &-\sqrt{5}&-\sqrt{5}&\sqrt{5} \end{array} \\\\\\\ [x-(-\sqrt{5})]^2+[y-(-\sqrt{5})]^2=(\sqrt{5})^2 \\\\\\ (x+\sqrt{5})^2+(y+\sqrt{5})^2=5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A~~%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter~%28%26%5Cquad%20%7B%7B%20h%7D%7D%2C%26%5Cquad%20%7B%7B%20k%7D%7D%5Cquad%20%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-%5Csqrt%7B5%7D%26-%5Csqrt%7B5%7D%26%5Csqrt%7B5%7D%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-%5Csqrt%7B5%7D%29%5D%5E2%2B%5By-%28-%5Csqrt%7B5%7D%29%5D%5E2%3D%28%5Csqrt%7B5%7D%29%5E2%0A%5C%5C%5C%5C%5C%5C%0A%28x%2B%5Csqrt%7B5%7D%29%5E2%2B%28y%2B%5Csqrt%7B5%7D%29%5E2%3D5)
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The area of a circle is represented by the equation:
We know that the diameter is two times the radius:
So if we know that the diameter of the circle is 24m, we can divide this by two in order to get the radius:
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So then we can plug this radius in to the equation for the area of a circle:
We are told to use 3.14 as pi, and not pi itself, so let's plug in 3.14 for pi:
Now we know that the area of this circle is 452.16 square meters.