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Oksanka [162]
3 years ago
10

What is the center of a circle represented by the equation (x-5)2+(y+6)2=42?

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
4 0

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-5)^2+(y+6)^2=42\implies [x-\stackrel{h}{5}]^2+[y-(\stackrel{k}{-6})]^2=(\stackrel{r}{\sqrt{42}})^2~~ \begin{cases} \stackrel{center}{(5,-6)}\\\\ \stackrel{radius}{\sqrt{42}} \end{cases}

kumpel [21]3 years ago
4 0

Answer:

<h2>(5, -6)</h2>

Step-by-step explanation:

The equation of a circle:

(x-h)^2+(y-k)^2=r^2

<em>(h, k)</em><em> - center</em>

<em>r</em><em> - radius</em>

<em />

We have:

(x-5)^2+(y+6)^2=42\\\\(x-5)^2+(y-(-6))^2=42

Therefore

h=5,\ k=-6,\ r^2=42\to r=\sqrt{42}

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