Question

Find an equation of the circle that satisfies the given conditions.

Center$\left(2,-1\right)$; radius 3

Short answer

The equation of the circle is ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$.

Step-by-step solution

Step 1. Apply the concept of circles.

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.

Step 2. Input the values of center and radius.

Consider the provided conditions that the center of a circle is at $\left(2,-1\right)$and radius is 3 units.

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, $\left(h,k\right)$and $\left(2,-1\right)$ also radius r is 3.

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

Substitute the values in the standard equation of a circle.

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

Step 3. Calculation step.

Simplify the standard equation of the circle obtained.

$\begin{array}{c}{\left(x-2\right)}^2+{\left(y-\left(-1\right)\right)}^2=3^2\\ {\left(x-2\right)}^2+{\left(y+1\right)}^2=9\end{array}$
Hence, the equation of a circle with center$\left(2,-1\right)$ and radius 3 is ${\left(x-2\right)}^2+{\left(y+1\right)}^2=9$.