Chapter P.2, Problem 42E

Solution

The standard form of equation for circle.

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

Given information:

Center of circle is: $\left(-3,2\right)$ , and radius of circle is 1.

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 standard form of equation for the given circle use the standard form of equation for circle:

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

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

Thus,

  $\begin{array}{l}h=-3,\\ k=2,\\ r=1\end{array}$

Thus, the standard form of equation for the given circle is:

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