Q 9. - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

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 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

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