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.!