Question

HELP ASAP !

b) findthe equation of circle having (2,3) and (4,0) as the end points of the diameter
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Answer 1

`\left( x-3\right)^(2) +\left( y-(3)/(2) \right)^(2) =(13)/(4)`

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

`r=(AB)/(2) =\frac{\sqrt{(0-3)^(2)+(4-2)^2} }{2} =(√(13) )/(2)`

Then Equation of the circle is:

`\left( x-3\right)^(2) +\left( y-(3)/(2) \right)^(2) = \left( (√(13) )/(2) \right)^(2) \\\Longrightarrow \left( x-3\right)^(2) +\left( y-(3)/(2) \right)^(2) =(13)/(4)`