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Questions: Algebra BusinessCalculus
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Find the center and radius of the circle \(\displaystyle \left(x - 10\right)^{2} + \left(y + 10\right)^{2}=49\).
Using the equation of a circle \(\displaystyle (x-h)^2+(y-k)^2=r^2\) to identify \(\displaystyle h,k,\) and \(\displaystyle r\) gives: \(\displaystyle h=10\), \(\displaystyle k=-10\), and \(\displaystyle r=7\). The center is at \(\displaystyle (10,-10)\) and the radius is \(\displaystyle r = 7\).
\begin{question}Find the center and radius of the circle $\left(x - 10\right)^{2} + \left(y + 10\right)^{2}=49$.
\soln{9cm}{Using the equation of a circle $(x-h)^2+(y-k)^2=r^2$ to identify $h,k,$ and $r$ gives: $h=10$, $k=-10$, and $r=7$. The center is at $(10,-10)$ and the radius is $r = 7$. }
\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 - 10\right)^{2} + \left(y + 10\right)^{2}=49 " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28x%20-%2010%5Cright%29%5E%7B2%7D%20%2B%20%5Cleft%28y%20%2B%2010%5Cright%29%5E%7B2%7D%3D49%20" alt="LaTeX: \displaystyle \left(x - 10\right)^{2} + \left(y + 10\right)^{2}=49 " data-equation-content=" \displaystyle \left(x - 10\right)^{2} + \left(y + 10\right)^{2}=49 " /> .</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=10 " src="/equation_images/%20%5Cdisplaystyle%20h%3D10%20" alt="LaTeX: \displaystyle h=10 " data-equation-content=" \displaystyle h=10 " /> , <img class="equation_image" title=" \displaystyle k=-10 " src="/equation_images/%20%5Cdisplaystyle%20k%3D-10%20" alt="LaTeX: \displaystyle k=-10 " data-equation-content=" \displaystyle k=-10 " /> , and <img class="equation_image" title=" \displaystyle r=7 " src="/equation_images/%20%5Cdisplaystyle%20r%3D7%20" alt="LaTeX: \displaystyle r=7 " data-equation-content=" \displaystyle r=7 " /> . The center is at <img class="equation_image" title=" \displaystyle (10,-10) " src="/equation_images/%20%5Cdisplaystyle%20%2810%2C-10%29%20" alt="LaTeX: \displaystyle (10,-10) " data-equation-content=" \displaystyle (10,-10) " /> and the radius is <img class="equation_image" title=" \displaystyle r = 7 " src="/equation_images/%20%5Cdisplaystyle%20r%20%3D%207%20" alt="LaTeX: \displaystyle r = 7 " data-equation-content=" \displaystyle r = 7 " /> . </p> </p>