Chapter P.2, Problem 45E

Solution

The center and radius of the circle.

The center of circle is $\left(3,1\right)$ and radius is 6.

Given information:

The equation of circle ${\left(x-3\right)}^2+{\left(y-1\right)}^2=36$ .

Formula used:

The standard form of equation for circle with center $\left(h,k\right)$ and radius of circle is r is given by

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

Calculation:

In order to find the center and radius of the given circle use the standard form of equation for the circle:

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

Here, $\left(h,k\right)$ represents the center of the circle and r is the radius of circle.

Note that the given equation of circle can be rewritten as:

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-1\right)}^2=36\\ {\left(x-3\right)}^2+{\left(y-1\right)}^2=6^2\end{array}$

Comparing it with standard form of equation of circle:

  $\begin{array}{l}h=3,\\ k=1,\\ r=6\end{array}$

Thus, the center of circle is $\left(3,1\right)$ and radius is 6.