Chapter C, Problem 1E

To determine

The equation of the circle with center $\left(3,-1\right)$ and radius $5$ .

Answer to Problem 1E

The equation of the circle is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=25$ .

Explanation of Solution

Given:

The center of the circle is $\left(3,-1\right)$ and radius $5$ .

Concept used:

The equation of the circle with center $\left(h,k\right)$ and radius $\left(r\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{....}\left(1\right)$

Calculation:

Since the center of the circle is $\left(3,-1\right)$ and radius $5$ . Therefore, from $\left(1\right)$ .

  $\begin{array}{c}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\mathrm{....}\left(\text{From }\left(1\right)\right)\\ {\left(x-3\right)}^2+{\left(y-\left(-1\right)\right)}^2={\left(5\right)}^2\\ {\left(x-3\right)}^2+{\left(y+1\right)}^2=25\end{array}$

Hence, the equation of the circle is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=25$ .