Problem 1
Answer: x^2 + (y - 13)^2 = 4
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Work Shown:
x^2 + y^2 - 26y + 165 = 0
x^2 + (y^2 - 26y) + 165 = 0
x^2 + (y^2 - 26y + 169) - 169 + 165 = 0 ... see note below
x^2 + (y - 13)^2 - 4 = 0
x^2 + (y - 13)^2 = 4
This is a circle with center (0,13) and radius 2
note: I took half of the y coefficient 26 to get 13, then I squared 13 to get 169. This is added and subtracted to help keep the equation balanced.
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Problem 2
Answer: (x+1)^2 + (y+12)^2 = 3
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Work Shown:
x^2 + y^2 + 2x + 24y + 142 = 0
(x^2 + 2x) + (y^2 + 24y) + 142 = 0
(x^2 + 2x + 1) - 1 + (y^2 + 24y + 144) - 144 + 142 = 0
(x+1)^2 + (y+12)^2 - 3 = 0
(x+1)^2 + (y+12)^2 = 3
This is a circle with center (-1,-12). The radius is sqrt(3).
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Problem 3
Answer: (x+1)^2 + (y+3)^2 = 169
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Work Shown:
Use the midpoint formula to find the center of the circle
xm = (x1+x2)/2
xm = (-1+(-1))/2
xm = -1
ym = (y1+y2)/2
ym = (-16+10)/2
ym = -3
The midpoint is (xm,ym) = (-1,-3)
The center of the circle is (h,k) = (-1,-3)
Use the distance formula to find the distance between the two given endpoints
d = sqrt((x1 - x2)^2 + (y1 - y2)^2)
d = sqrt((-1-(-1))^2 + (-16-10)^2)
d = sqrt((-1+1)^2 + (-16-10)^2)
d = sqrt((0)^2 + (-26)^2)
d = sqrt(0 + 676)
d = sqrt(676)
d = 26
The distance between the two endpoints is 26, so the diameter is 26
The radius is half that at r = d/2 = 26/2 = 13
Now we plug the center and radius into the standard equation of a circle
(x-h)^2 + (y-k)^2 = r^2
(x-(-1))^2 + (y-(-3))^2 = 13^2
(x+1)^2 + (y+3)^2 = 169
