Question 1019836: Find the equation of the circle touching the line 4x-3y=28 at (4,-4) and passing through (-3,-5).
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the equation of the circle touching the line 4x-3y=28 at (4,-4) and passing through (-3,-5).
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The center of the circle is on the line perpendicular to the given line thru (4,-4).
The slope of the line is -3/4 --> y+4 = (-3/4)*(x-4)
--> y = -3x/4 - 1
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The slope of a line tangent to a circle is -x/y.
The center of the circle is also on the line perpendicular to the tangent line thru (-3,-5).
The slope of the tangent line = -3/5.
The slope of the line perpendicular = 5/3
--> y+5 = (5/3)*(x+3)
--> y = 5x/3
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The center of the circle is the intersection of y = 5x/3 and y = -3x/4 - 1
5x/3 = -3x/4 - 1
20x = -9x - 12
x = -12/29 --> y = -20/29
Center at (-12/29,-20/29)
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r = distance from the center to (4,-4)
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Something is wrong, I'll check it later.
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I can't find the error. If you can't find it, repost it, maybe someone else will do it.
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I see what the problem is. A slope of -x/y for a circle applies to circles centered at the Origin.
I'll work on this later today.
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