equation of line tangent to circle
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A short worked problem dated 9.10.09 and signed PhL, so it is Phil's own note. It derives the slope and intercept of the tangent line to a circle of radius r at angle θ by two methods: one from the condition that the tangent is perpendicular to the radius vector, the other from the normalized normal vector of the line. Both give y = -cotθ x + r/sinθ. The diagram did not come through in the text.
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Problem: What is the equation of a line tangent to a circle? PhL 9.10.09
First, here is the picture:
Method 1. What are m and b in terms of r and θ ? Start with:
(r-r1)r1 = 0 // line is tangent to the circle at point r1
Write out the first equation to get
[ ( x,y) - (x1, y1) ] (x1, y1) = 0
(x-x1)x1 + (y-y1)y1 = 0 => (x-x1)x1 = – (y-y1)y1
–(x-x1)x1/y1 = (y-y1) => y = y1 –(x1/y1)x +x12/y1 = mx+b
Therefore
m = –(x1/y1) = -cotθ
b = y1 + x12/y1 =(y12 + x12)/y1 = r2/y1 = r/sinθ
So the line tangent to a circle of radius r at location θ is given by
y = [-cotθ ] x + r/sinθ
Method 2.
mx-y-b = 0 => n = (m,-1) // unnormalized normal vector to line
= (m,-1)/ = 1 = r1/r = (x1, y1)/r
Thus we have two equations
m/ = x1/r = cosθ
-1/ = y1/r = sinθ
Divide these equations to get
-m = cotθ
Now insist that r1 is on the line y = mx+b so that
y1 = [-cotθ] x1 + b
rsinθ = [-cotθ]rcosθ + b = - r cos2θ/sinθ + b
Then
b = rsinθ + r cos2θ/sinθ = r( sin2θ + cos2θ)/sinθ = r/sinθ
and we get the same result as by Method 1.