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The endpoints of a diameter of a circle are  (6,2) and (−2,5) .

What is the standard form of the equation of this circle?

 Jan 17, 2020
 #1
avatar+22108 
+3

Find the diameter     sqrt (x1-x2)^2  + (y1-y2)^2  )       1/2 of this is r    the radius

 

Find the MIDPOINT ....this will be the center   (h,k)

    from 6 to -2 is  8 units     midpoint will be 4 units from either end    at       2

    from 2 to 5   is 3 units      midpoint will be  1 1/2 from either end  at      3 1/2      so (h.k) = (2, 3 1/2)

 

Standard form of a circle is

 

(x-h)^2 + (y-k)^2 = r^2       You now have all of the building parts....put it together and    done.

 Jan 17, 2020
 #2
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+1

Hi, can you please set up the diameter equation for me so I can solve it because I think I keep doing it wrong somehow

Guest Jan 17, 2020
 #3
avatar+22108 
+2

diameter =   sqrt   ( 6- -2)^2  + (2-5)^2   )   

                = sqrt (8^2 + (-3)^2 )

                = sqrt (64 + 9 )

                  = sqrt 73

radius = sqrt 73   /2         r^2 =  73/4

ElectricPavlov  Jan 17, 2020

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