Chapter 1.8, Problem 89E

To determine

To calculate: Thecenter and radius of the circle ${\left(x-3\right)}^2+y^2=16$ and sketch the graph of the circle.

Answer to Problem 89E

The center of the circle is $\left(3,0\right)$ and radius is $4$ units. The graph of the circle is provided below,

  

Explanation of Solution

Given information:

The equation ${\left(x-3\right)}^2+y^2=16$ .

Formula used:

The standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Calculation:

Consider the equation ${\left(x-3\right)}^2+y^2=16$ .

Rewrite the equation as,

  ${\left(x-3\right)}^2+{\left(y-0\right)}^2={\left(4\right)}^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ and ${\left(x-3\right)}^2+{\left(y-0\right)}^2={\left(4\right)}^2$ .

Here, $h=3,k=0\text{ and }r=4$ .

Therefore, center of circle is $\left(3,0\right)$ and radius is $4$ .

Now, on the Cartesian plane construct a circle with center at $\left(3,0\right)$ and radius $4$ units.

The graph obtained is provided below,

  

Thus, the center of the circle is $\left(3,0\right)$ and radius is $4$ units.