How do you find the center of the circle that can be circumscribed about ΔEFG with E (2,2), F (2,-2), and G (6,-2)?

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Answer 1 · 2016-09-14T16:37:52

#" The circumcentre is "(4,0)#.

Observe that the pts. #E(2,2) and F(2,-2)# have the same x-co-

ordinates.

#":. line "EF" is vertical", i.e., ||" to "Y"-axis"#.

Similarly, #"line "FG" is horizontal, i.e.,"||" to "X"-axis"#.

Therefore, #m/_EFG=90^@, &, "in "DeltaEFG, EG" is hypotenuse."#

We know from Geometry that, in a right-#Delta#, the mid-pt. of the

hypotenuse is the circumcentre of the #Delta#.

#:." The circumcentre is the mid-pt. of seg. "EG=((2+6)/2, (2-2)/2)#

#=(4,0)#.

Enjoy Maths.!