Solution
- The diameter of a circle is the longest chord of a circle.
- Thus, the coordinate of the center of the circle is the midpoint of the diameter or longest chord.
![\begin{gathered} \text{ The formula for finding the Midpoint is:} \\ M(x,y)=((x_2+x_1)/(2),(y_2+y_1)/(2)) \\ \\ (x_1,y_1)=(4,5.5) \\ (x_2,y_2)=(4,10.5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rx5d53o3s8whie9dcuup1dr8licmv2ei3g.png)
- Thus, we can solve the question as follows:
![\begin{gathered} M=(4+4)/(2),(10.5+5.5)/(2) \\ \\ M=(4,8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fg239zojrohpe8kns94t1gpxecjxs1i5os.png)
The center of the circle is (4, 8)
- The radius can be gotten by finding the distance between the coordinate of the center and any of the endpoints of the diameter.
- Thus, we have:
![\begin{gathered} D=√((x_2-x_1)^2+(y_2-y_1)^2) \\ \\ r=√((4-4)^2+(8-5.5)^2) \\ \\ r=√(2.5^2) \\ \\ r=2.5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vgnx156s3t2hzb0h5u28x09y9zcdm03rpu.png)
- The radius has a magnitude of 2.5 units
- The equation of the circle can be gotten by the formula given below:
![\begin{gathered} \text{ Equation of a circle} \\ r^2=(x-a)^2+(y-b)^2 \\ where, \\ (a,b)\text{ is the center of the circle} \\ r\text{ is the radius} \\ \\ 2.5^2=(x-4)^2+(y-8)^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/36v214ik09pjdiycc5lph5u3atgc5hqzsb.png)
The equation of the circle becomes:
![(x-4)^2+(y-8)^2=2.5^2](https://img.qammunity.org/2023/formulas/mathematics/college/c262ucrf79jwqafa27mjzjcgh7rohh3fho.png)