Question

Find the center and radius of the circle. Write the standard form of the equation.

Short answer

Center is $\left(\frac{5}{2},2\right)$.

Radius is $\frac{3}{2}$units.

Standard form of the equation is${\left(x-\frac{5}{2}\right)}^2+(y-2{)}^2=\frac{9}{4}$.

Step-by-step solution

Step 1. Given information

Figure of a circle is as follows:

Step 2. Find the center and radius of a circle.

Diameter is the distance between the points $\left(1,2\right)$and $\left(4,2\right)$

Diameter$=\sqrt{(4-1{)}^2+(2-2{)}^2}$

$$\begin{gathered} =\sqrt{9+0} \\ =3 \end{gathered}$$

$\begin{array}{rcl}\text{Radius}& =& \frac{\text{Diameter}}{2}\\ & =& \frac{3}{2}\end{array}$

Center is the midpoint of the points $\left(1,2\right)$and $\left(4,2\right)$

$\begin{array}{rcl}\text{Center}& =& \left(\frac{1+4}{2},\frac{2+2}{2}\right)\\ & =& \left(\frac{5}{2},2\right)\end{array}$

Step 3. Find the standard form of the equation of a circle.

Equation of a circle with center $\left(h,k\right)$and radius $r$is $(x-h{)}^2+(y-k{)}^2=r^2$.

Therefore, equation of the circle with center $\left(\frac{5}{2},2\right)$and radius $\frac{3}{2}$is

role="math" localid="1646125761348" $\begin{array}{rcl}{\left(x-\frac{5}{2}\right)}^2+(y-2{)}^2& =& {\left(\frac{3}{2}\right)}^2\\ {\left(x-\frac{5}{2}\right)}^2+(y-2{)}^2& =& \frac{9}{4}\end{array}$