Chapter 1.8, Problem 6E

To determine

To fill: The blank in the statement “The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center $\left(,\right)$ and radius $$ .”

Answer to Problem 6E

The complete statement is “The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center $\left(1,2\right)$ and radius $3$units.”

Explanation of Solution

Given information:

The statement “The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center $\left(,\right)$ and radius $$ .”

Consider the provided statement “The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center $\left(,\right)$ and radius $$ .”

Rewrite the equation as ${\left(x-1\right)}^2+{\left(y-2\right)}^2=3^2$

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 ${\left(x-1\right)}^2+{\left(y-2\right)}^2=3^2$ .

Here, $h=1,k=2\text{ and }r=3$ .

Therefore, center of circle is $\left(1,2\right)$ and radius is $3$ .

Thus, the complete statement is “The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center $\left(1,2\right)$ and radius $3$units.”