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Questions: Algebra BusinessCalculus
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Find the center and radius of the circle \(\displaystyle \left(x + 6\right)^{2} + \left(y + 8\right)^{2}=64\).
Using the equation of a circle \(\displaystyle (x-h)^2+(y-k)^2=r^2\) to identify \(\displaystyle h,k,\) and \(\displaystyle r\) gives: \(\displaystyle h=-6\), \(\displaystyle k=-8\), and \(\displaystyle r=8\). The center is at \(\displaystyle (-6,-8)\) and the radius is \(\displaystyle r = 8\).
\begin{question}Find the center and radius of the circle $\left(x + 6\right)^{2} + \left(y + 8\right)^{2}=64$.
\soln{9cm}{Using the equation of a circle $(x-h)^2+(y-k)^2=r^2$ to identify $h,k,$ and $r$ gives: $h=-6$, $k=-8$, and $r=8$. The center is at $(-6,-8)$ and the radius is $r = 8$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the center and radius of the circle <img class="equation_image" title=" \displaystyle \left(x + 6\right)^{2} + \left(y + 8\right)^{2}=64 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20%2B%206%5Cright%29%5E%7B2%7D%20%2B%20%5Cleft%28y%20%2B%208%5Cright%29%5E%7B2%7D%3D64%20" alt="LaTeX: \displaystyle \left(x + 6\right)^{2} + \left(y + 8\right)^{2}=64 " data-equation-content=" \displaystyle \left(x + 6\right)^{2} + \left(y + 8\right)^{2}=64 " /> .</p> </p><p> <p>Using the equation of a circle <img class="equation_image" title=" \displaystyle (x-h)^2+(y-k)^2=r^2 " src="/equation_images/%20%5Cdisplaystyle%20%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2%20" alt="LaTeX: \displaystyle (x-h)^2+(y-k)^2=r^2 " data-equation-content=" \displaystyle (x-h)^2+(y-k)^2=r^2 " /> to identify <img class="equation_image" title=" \displaystyle h,k, " src="/equation_images/%20%5Cdisplaystyle%20h%2Ck%2C%20" alt="LaTeX: \displaystyle h,k, " data-equation-content=" \displaystyle h,k, " /> and <img class="equation_image" title=" \displaystyle r " src="/equation_images/%20%5Cdisplaystyle%20r%20" alt="LaTeX: \displaystyle r " data-equation-content=" \displaystyle r " /> gives: <img class="equation_image" title=" \displaystyle h=-6 " src="/equation_images/%20%5Cdisplaystyle%20h%3D-6%20" alt="LaTeX: \displaystyle h=-6 " data-equation-content=" \displaystyle h=-6 " /> , <img class="equation_image" title=" \displaystyle k=-8 " src="/equation_images/%20%5Cdisplaystyle%20k%3D-8%20" alt="LaTeX: \displaystyle k=-8 " data-equation-content=" \displaystyle k=-8 " /> , and <img class="equation_image" title=" \displaystyle r=8 " src="/equation_images/%20%5Cdisplaystyle%20r%3D8%20" alt="LaTeX: \displaystyle r=8 " data-equation-content=" \displaystyle r=8 " /> . The center is at <img class="equation_image" title=" \displaystyle (-6,-8) " src="/equation_images/%20%5Cdisplaystyle%20%28-6%2C-8%29%20" alt="LaTeX: \displaystyle (-6,-8) " data-equation-content=" \displaystyle (-6,-8) " /> and the radius is <img class="equation_image" title=" \displaystyle r = 8 " src="/equation_images/%20%5Cdisplaystyle%20r%20%3D%208%20" alt="LaTeX: \displaystyle r = 8 " data-equation-content=" \displaystyle r = 8 " /> . </p> </p>