Question 995162: Help me find the equation of a circle passing through points (6,0) and (24,0) and is a tangent to the y-axis
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Help me find the equation of a circle passing through points (6,0) and (24,0) and is a tangent to the y-axis
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Sketch the figure.
The perpendicular bisector of the segment joining (6,0) and (24,0)
is the line x = 6+(24-6)/2 = 15
So the center is at (15,k).
Therefore the radius = 15, since the circle is tangent to the y-axis.
Then:
(x-15)^2 + (y-k)^2 = 15^2
Using (6,0) you get:
(81)+(0-k)^2 = 225
k^2 = 144
k = +12 or -12
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Ans: (x-15)^2 + (y-12)^2 = 15^2
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Cheers,
Stan H.
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Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Help me find the equation of a circle passing through points (6,0) and (24,0) and is a tangent to the y-axis
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Label the points A(6,0) and B(24,0)
The center of the circle, point C, is on the perpendicular bisector of AB, which is x = 15
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Tangent to the y-axis makes the radius 15.
The distance from A and B to the center is also 15.
y = 12
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Center at (15,12) and (15,-12) (there are 2 circles)
(15,12) --> 
(15,-12) --> 
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