Chapter 9.2, Problem 3OE

To determine

To calculate: The center and radius of the circle.

Answer to Problem 3OE

The center and radius of the circle are $\left(2,4\right)$ and $6$ respectively.

Explanation of Solution

Given information: The equation of the circle is ${\left(x-2\right)}^2+{\left(y-4\right)}^2=36$ .

Formula used: ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Calculation: Comparing ${\left(x-2\right)}^2+{\left(y-4\right)}^2=36$ to ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , we get,

  $\begin{array}{ccc}\begin{array}{l}-h=-2\\ \therefore h=2\end{array}& \begin{array}{l}-k=-4\\ \therefore k=4\end{array}& \begin{array}{l}{r}^2=36\\ \therefore r=6\end{array}\end{array}$

Here, the radius of the circle is 6.

Now, as we know that $\left(h,k\right)$ is the center of a circle.

Therefore, the center of the given circle $\left(h,k\right)=\left(2,4\right)$

Hence, the required center and radius of the circle are $\left(2,4\right)$ and $6$ respectively.