Question

The standard equation of a circle with center at $(2,-3)$and radius 1 is ________.

Short answer

The standard equation of a circle with center at $(2,-3)$and radius 1 is${\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{(}\mathit{y}\mathbf{+}\mathbf{3}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{1}$.

Step-by-step solution

Step 1. Given Information

We want to find the equation of a circle with centre in$(2,-3)$and radius 1.

Step 2. Graph the circle:

Looking at the graph, to find the equation,

Let $(x,y)$represent the coordinates of any point on circle that has a radius $r$and the center in the point $(h,k)$.

Then the distance between the points $(x,y)and(h,k)$is calculated using the standard formula for distance between two points$\sqrt{{(x-h)}^2+{(y-k)}^2}$

Step 3. Find equation of circle

In order for a point to be on the circle with a radius of $r$the distance between the given point $(x,y)$and the center of the circle has to be equal to $r$.

Therefore

role="math" localid="1646801546199" $$\begin{gathered} \sqrt{{(x-h)}^2+{(y-k)}^2}=r \\ {(x-h)}^2+{(y-k)}^2=r^2 \end{gathered}$$

which is the standard form of an equation of a circle with radius $r$and center $(h,k)$.

Now, since in the exercise role="math" localid="1646801699878" $(h,k)=(2,-3)andr=1$to obtain equation of a circle with the center in $(2,-3)$and with a radius 1 plug in role="math" localid="1646801794084" $h=2,k=-3andr=1$:

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