Question

find the center of the circle that you can circumscriBe about EFG with E(6,2), F(6,-2), and G(10,-2) EQUALS

Answer 1

Since the points form a right-triangle, the centre is at the mid-point between the first and the third points, namely ((6+10)/2, (2+(-2))/2)=(8,0)

Answer 2

can you explain this problem more detailed pretty please?

Answer 3

I suppose you are doing lines in analytic geometry, with slopes, and centres of lines etc.

The centre of a circle is typically found by the intersection of two perpendicular bisectors, say EF and FG.

It turns out that that EF is a vertical line, so the perpendicular bisector is the horizontal line y=0.
Also, FG is a horizontal line, so the perpendicular bisector is x=8.
So we conclude that the centre is at (8,0) without much ado.

In general, a few calculations are required, typically, find the equation of two lines, the mid points of the chords, and lines perpendicular to these chords, and finally find the intersection. A little snooping saved us all this trouble.

Answer 4

THANK YOU SO MUCH

Answer 5

Your welcome!