Posted by glysdi on Friday, September 2, 2011 at 2:14am.
a circle is centered at (3,4) and in tangent to 5x4y+12=0.find its equation.

analytic and solid geometry  Reiny, Friday, September 2, 2011 at 8:19am
equation must be
(x+3)^2 + (y4)^2 = k , were √k is the radius
2(x+3) + 2(y4)dy/dx = 0
dy/dx = (x+3)/(y4) which is also the slope of the tangent.
But from 5x  4y + 12 = 0 , the slope of the tangent is 5/4
so (x3)/(y4) = 5/4
5y  20 = 4x  12
4x + 5y = 8
4x+5y=8 times 5 > 20x + 25y = 40
5x4y= 12 times 4 > 20x  16y = 48
subtract: 41y = 82
y = 2
then in 4x+5y = 8
4x + 10 = 8
x = 1/2 , the point of contact is (1/2, 2) or (.5,2)
so back in original:
(1/2 + 3)^2 + (24)^2 = k
k = 41/4 or 10.25
(x+3)^2 + (y4)^2 = 41/4
check:
at (.5,2) dy/dx = (.5+3)/(24) = 2.5/2 = 5/4
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