Chapter 9.3, Problem 6MRE

To determine

To find: The equation of the circle.

Answer to Problem 6MRE

The required equation is ${\left(x+1\right)}^2+{\left(y-3\right)}^2=16$ .

Explanation of Solution

Given information: the coordinate of center is $\left(-1,3\right)$ and the length of the radius is $4$ .

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

Calculation: As we know that the circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ has the center $\left(h,k\right)$ and radius $\left(r\right)$ .

So, $\left(h,k\right)=\left(-1,3\right)$ and $r=4$

Now substitute these values into ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ to get the equation of the circle.

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ {\left(x-\left(-1\right)\right)}^2+{\left(y-3\right)}^2=4^2\\ {\left(x+1\right)}^2+{\left(y-3\right)}^2=16\end{array}$

Therefore, the required equation is ${\left(x+1\right)}^2+{\left(y-3\right)}^2=16$ .