Circle Proof

Can anyone prove that the tangent to a circle at a given point is perpendicular to the radius, without using calculus? I can't immediately see how...

From http://www.hyperdictionary.com/dictionary/tangent

Quote:

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a straight line or plane that touches a curve or curved surface at a point but does not intersect it at that point

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Hmmm.

lets see.

If you've got a line that is Perpindicular to the radius at a given point on a circle, Plus some delta angle, thru trig you should be able to prove that the line must intersect some other point on the circle, for all delta angles <> 0. So, if two points on a circle satisfies this line, then by definition the line is not a tangent.

Only when the Delta angle is zero would only 1 point on the circle satisfy that line. Which would make it a tangent. And if the Delta angle is zero, then it is perpindicular.

I'm not up to pulling the actual mathematical analysis together at this moment, but I think this approach would be valid.


:wave:
-Lou

Not too hard:
1) You have a tangent meeting the circle at point P
2) Draw a diameter from P through the centre to point Q
3) Choose any point R elsewhere in the circle (not at P or Q)
4) Join P, Q, R....you get rightangled triangle..correct? (angle PRQ subtended by diameter)
5) Now you know the radius must be at right angles to the tangent, due to the Alternate Segment theorem.

Tada! No calculus, or even limits!

I don't think you can do that because the alternate segment thing that I know of is only true because the radius is perpendicular to the tangent. So illustrate it or something...

Also Lou I can't visualise what you're trying to do, what is this delta-angle?

Lets say you have some angle A = 90 + b

Where b can be from 0 up to 360 degrees.

b is what I refer to as delta angle.

So if you say some point {X₁, Y₁}
is on the circle, and you draw a radius to that point, then build a fromula, using trig, for a line that uses that point, where the line is at 90 + b degrees from the radius angle. Using the formula for a circle, show the line intersects the circle at 2 points for all angles b, except when b is = 0.

Which shows that when b = 0, there exists only 1 point on the circle that satisfies that line. Which means the line is a tangent to the circle. And since that line is a 90 + b degres from the radius, or 90 {since b = 0} then the line is at right angles to the radius or circle.

;)
-Lou

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I don't think you can do that because the alternate segment thing that I know of is only true because the radius is perpendicular to the tangent.

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Are u sure? Can it be that the radius is perpendicular to the tangent because of the alternate segment theorem? :p


P.S. it's not really calculus, just uses simple logical limits, and you can think of it as this:
1) Start with circle centre X, and a point (at 12 o'clock) Y
2) Now, have a point on the circle somewhere to the left (L), and a point on the circle somewhere to the right (R)
3) As you move L towards Y, the line gets flatter and flatter, but still from bottom-left to top-right.
4) As you move R towards Y, the line gets flatter and flatter, but still from bottom-right to top-left.
5) Combining (3) and (4), the line must be flat at some point, which can't be when L isn't Y, and can't be when R isn't Y (cos then these wouldn't be flat)
=> It must be when L = R = Y, at the tangent!

I would have said that was also beating around the bush, since you only know the line is flat because it's perpendicular to another line (the y-axis which just happens to be the radius). Try putting Y at a random point on the circle and deriving it then.

Use intersection theorem to find intersection (X, Y) of radius and line.

Then...

(X^2) + (Y^2) = Radius^2

If given point on line is endpoint of radius then

cX = center circle x
cY = " " Y

Find slope of (cX, cY) & (X, Y)

lX# = Line endpoint X#
lY# = Line endpoint Y#

Find slope of (lX1, lY1) & (lX2, lY2)

If Line slope = -1/m of radius then by definition, line is tangent to radius.

Simple algebra. ;)

Consider the circle to be one point (because you could derive all other points by rotating that point around the circle.) Let our point be (0, r) where r is our radius.

We know that the tangent line to this point is y = r. If we consider the radius as a line, it would be x = 0. These two lines are perpendicular by definition. Therefore, all tangents are perpendicular to their respective radii.

In order to get a formal proof of this, you will have to use calculus.