[tex]\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}[/tex]
Answer:
Step-by-step explanation:
The equation of a circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
We have:
[tex](x-5)^2+(y+6)^2=42\\\\(x-5)^2+(y-(-6))^2=42[/tex]
Therefore
[tex]h=5,\ k=-6,\ r^2=42\to r=\sqrt{42}[/tex]
