Asked by Peluche 🐻
                There are three points at the rim of the circle: (-4,1) (-2,-3) (5,-2). Where is the center of the lake? What is it's diameter if each unit of a coordinate plane represents 3/5 of a mile?
            
            
        Answers
                    Answered by
            Reiny
            
    midpoint for (-4,1) and (-2,-3)
= (-3,-1)
slope through the two given points
= -4/2 = -2
so slope of perpendicular = 1/2
equation of right-bisector:
y+1 = (1/2)(x+3)
2y + 2= x+3
x- 2y = -1 **
midpoint of (-4,1) and (5,-2) is (1/2 , -1/2)
slope of chord = -3/9 = -1/3
so slope of perpendicular = 3
equation of right bisector:
y + 1/2 = 3(x-1/2)
y + 1/2 = 3x - 3/2
2y + 1 = 6x - 3
6x - 2y = 4 ***
subtract equations ** and ***
5x = 5
x = 1
then in **
1-2y = -1
-2y = -2
y = 1
the centre is (1,1) , and the equation is
(x-1)^2 + (y-1)^2 = r^2
sub in (-4,1)
25 + 0 = r^2
equation: (x-1)^2 + (y-1)^2 = 25
so the radius is 5 and the diameter is 10 units on the grid, or
10(3/5) miles = 6 miles
    
= (-3,-1)
slope through the two given points
= -4/2 = -2
so slope of perpendicular = 1/2
equation of right-bisector:
y+1 = (1/2)(x+3)
2y + 2= x+3
x- 2y = -1 **
midpoint of (-4,1) and (5,-2) is (1/2 , -1/2)
slope of chord = -3/9 = -1/3
so slope of perpendicular = 3
equation of right bisector:
y + 1/2 = 3(x-1/2)
y + 1/2 = 3x - 3/2
2y + 1 = 6x - 3
6x - 2y = 4 ***
subtract equations ** and ***
5x = 5
x = 1
then in **
1-2y = -1
-2y = -2
y = 1
the centre is (1,1) , and the equation is
(x-1)^2 + (y-1)^2 = r^2
sub in (-4,1)
25 + 0 = r^2
equation: (x-1)^2 + (y-1)^2 = 25
so the radius is 5 and the diameter is 10 units on the grid, or
10(3/5) miles = 6 miles
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