Calculating tangents on a circle given r, θ, and (h,k), without using Calculus.
Figure 1 shows a circle with a line (lt) tangent.
Consider the line in figure 2. The center of the circle is (0,0), so if we are to describe the line passing through both the center of the circle, and the arc of the angle θ, using y=mx+b, we can remove the b (y-intercept = 0), and conclude that y=mx is sufficient in this situation.
Consider m, slope, = ∆y/∆x. Any two non-identical points on a line can be used to calculate slope. In the special case of a line passing through (0,0), slope is simply any current y value, over a corresponding x value.
The trig functions sin and cosine (or simply “cos”) will translate from angle to point, so sin(θ)/cos(θ) = m for any line passing through the center of a circle at (0,0), with a given angle (θ). Sine over cosine has a dedicated function called “tangent”, or simply “tan”. Therefore, the line in figure 2 has a formula of: y=tan(θ)x.
Consider the tangent lines in figures 3 and 4:
They share only one point, the point of tangency, with the radial line. (P1, which has different coordinates per angle.) Remember, a point is half of what is needed to describe a line.
Notice also, that as θ changes (and the radial line “rotates”), the tangent line moves about the circle, always making a 90-degree angle with the radial line. Just making a quick visual comparison between the radial line and the tangent line, you can see that the tangent line’s slope is negative, while the radial line has a positive slope. This relationship holds throughout these models, although for some angles the opposite will be true.
The lines in figure 5 have slopes with additive inverses. But you can tell they are not restricted to a 90-degree relationship, like the radius and tangent lines are.
The answer is that the slopes are not only negative, but also inverted. If our radial line has a slope of 1/2, the tangent line must have a slope of -2. So the slope of the tangent line = -1/tan(θ). (Otherwise known as -1•cotan(θ), but its easier to see these relationships with the former.)
The lines in figure 6 have the following formulas:
line-r: y=tan(θ)x
line-u: y=(-1/tan(θ))x
Now all we have to do is move the would-be tangent line (u) into position by an amount equal to the radius (in this example, 1) in a direction indicated by θ. The distance on the x-axis of this will be the y-intercept (b).
Consider figure 7:
The tangent line forms a right-triangle with the y-intercept, the point of tangency, and the center of the circle.
Right triangles have very special properties, but a little common sense will tell you that knowing a side, and two angles (90-degrees and θ-prime, which is just 90-θ), will identify one triangle from any other triangle.
For any right-triangle, the cosine of an angle times the length of the hypotenuse (”b” in our case) equals the length of the non-hypotenuse leg (r) that connects with the given angle. Re-writing and substituting, we get b=1/cos(90-θ). cos(90-θ) equals sin(θ) and 1/sin(θ) equals cosecant(θ), or csc(θ).
This yields the universal formula:
y=-1cotan(θ)x + csc(θ)
Larger radii can easily be accommodated by changing csc(θ) to r/sin(θ). Furthermore, the center of the circle (h,k) may be inserted into the formula as follows:
y=-1cotan(θ)(x-h)+(r/sin(θ))+k
