Chapter 11.3, Problem 83STP

To determine

Find the center and radius of the circle and graph the circle.

Answer to Problem 83STP

Center (3, − 7) and radius is $5\sqrt{2}$

Explanation of Solution

Given:

The equation of circle is ${(x-3)}^2+{(y+7)}^2=50$

Concept Used:

The general equation of circle is ${(x-h)}^2+{(y-k)}^2=r^2$ , where $\text{center}(h,k)\text{and}\text{radius}=r$

Calculation:

Comparing both equations: ${(x-h)}^2+{(y-k)}^2=r^2$ and ${(x-3)}^2+{(y+7)}^2=50$

$\begin{array}{}$ {(x-h)}^2+{(y-k)}^2=r^2$center$ \left(h_{1}k\right)$\text{and radius}$ =r$$ {(x-3)}^2+{(y+7)}^2=50$$ h=3;k=-7;r=\sqrt{50}=5\sqrt{2}$\end{array}$ Center (3, − 7) and radius is $5\sqrt{2}$

The graph of the circle ${(x-3)}^2+{(y+7)}^2=50$

Thus, Center (3, − 7) and radius is $5\sqrt{2}$