Find the center of the circle that can be circumscribed about efg with e(6, 4), f(6, 2), and g(10, 2).

 The general form of the equation of a circle is 

(x-a)^2 + (y-b)^2 = r^2   where the center is (a,b) 
Plugging in the given values:-

(6 - a)^2 + (4-b)^2 = r^2    and
(6-a)^2 + (2-b)^2  = r^2

Subtracting:-

(4-b)^2 - (2 - b)^2  = 0

16 - 8b + b^ - (4 -4b + b^2) = 0

12 - 4b  = 0
-4b =  -12

b = 3 

Similarly if we take the 2 points (6,2) and G(10,2)  and make two equations we can find the value of a

(6-a)^2 + (2-b)^2  = r^2
(10.a)^2 + (2-b)^2 = r^2    subtract::-

(6-a)^2 -(10-a(^2 = 0

this will give a = 8

so the center of the circle is  the poit (8,3)

The answer you are looking for is right in front of you it’s c. 3/20