Answer:
[tex]\left( x-3\right)^{2} +\left( y-\frac{3}{2} \right)^{2} =\frac{13}{4}[/tex]
Step-by-step explanation:
Let A(2,3) and B(4,0) be the end points of the diameter
then the segment AB is the diameter of the circle
Then the center of the circle (point M) is the midpoint of segment AB
Then M( (2+4)/2 , (3+0)/2 ) then M( 3 , 3/2 )
Now let’s find the radius r of our circle :
r = AB/2
Then
[tex]r=\frac{AB}{2} =\frac{\sqrt{(0-3)^{2}+(4-2)^2} }{2} =\frac{\sqrt{13} }{2}[/tex]
Then Equation of the circle is:
[tex]\left( x-3\right)^{2} +\left( y-\frac{3}{2} \right)^{2} = \left( \frac{\sqrt{13} }{2} \right)^{2} \\\Longrightarrow \left( x-3\right)^{2} +\left( y-\frac{3}{2} \right)^{2} =\frac{13}{4}[/tex]
