Points (–9, 2) and (–5, 6) are endpoints of the diameter of a circle What is the length of the diameter? What is the center point C of the circle? Given the point C you found in part (b), state the point symmetric to C about the x-axis

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Answer 1 · 2018-05-24T04:32:11

#d = sqrt(32) = 4sqrt(2) ~~ 5.66#
center, #C = (-7, 4)#
symmetric point about #x#-axis: #(-7, -4)#

Given: endpoints of the diameter of a circle: #(-9, 2), (-5, 6)#

Use the distance formula to find the length of the diameter: #d = sqrt((y_2 - y_1)^2 + (x_2 - x_1)^2)#

#d = sqrt((-9 - -5)^2 + (2 - 6)^2) = sqrt(16 + 16) = sqrt(32) = sqrt(16)sqrt(2) = 4 sqrt(2) ~~ 5.66#

Use the midpoint formula to find the center: #((x_1 + x_2)/2, (y_1 + y_1)/2)#:

#C = ((-9 + -5)/2, (2 + 6)/2) = (-14/2, 8/2) = (-7, 4)#

Use the coordinate rule for reflection about the #x#-axis #(x, y) -> (x, -y)#:

#(-7, 4)# symmetric point about #x#-axis: #(-7, -4)#

Answer 2 · 2018-05-24T04:34:40

1) #4 sqrt(2)# units.
2) #(-7,4)#
3) #(7,4)#

Let the point A be #(-9,2)# & Let the point B be #(-5,6)#

As the points #A# and #B# be the endpoints of the diameter of the circle. Hence, the distance #AB# be length of the diameter.

Length of the diameter#= sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Length of the diameter#= sqrt((-5+9)^2+(6-2)^2)#

Length of the diameter#= sqrt((4)^2+(4)^2)#

Length of the diameter#= sqrt(32)#

Length of the diameter#=4 sqrt(2)# units.

The centre of the circle is the midpoints of the endpoints of the diameter.

So, by midpoints formula,

#x_0= (x_1+x_2)/2# & #y_0= (y_1+y_2)/2#

#x_0= (-9-5)/2# & #y_0= (2+6)/2#

#x_0= (-14)/2# & #y_0= (8)/2#

#x_0= -7# & #y_0= 4#

Co-ordinates of the centre#(C) #= #(-7,4)#

The point symmetric to C about the x-axis has co-ordinates =#(7,4)#