Question

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

Answer 1

(5, -6)

Explanation:

The equation of a circle:

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

(h, k) - center

r - radius

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=√(42)`

Answer 2

`\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}{√(42)})^2~~ \begin{cases} \stackrel{center}{(5,-6)}\\\\ \stackrel{radius}{√(42)} \end{cases}`