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Question 331802: What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0.
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Find the bisector of the two tangent lines, x-3y-11=0 and 3x-y-9=0.
To do that, find the intersection of the 2 lines.
The slope of the bisector is tan(arctan(1 line) + arctan(other line))
The slope = 1
With the slope and the point of intersection, find the equation of the bisector.
Then find the intersection of the bisector and the line x+2y+19 = 0. That's the center of the circle.
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Find the distance from the center to either x-3y-11=0 or 3x-y-9=0, that's the radius.
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email for any assistance needed at moralloophole@aol.com
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