Proof That Two Tangents Can Be Drawn from a Point Outside a Circle

Question

I am studying the properties of circles and came across a standard geometric principle that states two tangents can be drawn from any external point to a given circle. While I understand the principle intuitively, I would like to delve deeper into the geometric proof of this concept. Here is my current understanding and approach:

• Concept and Definitions:

  • Let circle ( O ) have a radius ( r ), and let ( P ) be a point outside the circle such that the distance from ( P ) to the center ( O ) is greater than ( r ).

• Proof Approach:

  • Draw line segment ( OP ) where ( O ) is the center of the circle and ( P ) is the external point.
  • Find point ( M ) on ( OP ) such that ( $PM = r$ ) (thus ( M ) is outside the circle as ( P ) is outside and ( $PM \neq r$ ) of the circle).
  • Draw a circle centered at ( M ) with radius ( PM ). This circle intersects circle ( O ) at two points, say ( A ) and ( B ).
  • By definition, segments ( PA ) and ( PB ) are radii of the new circle and equal in length, and by the circle's property, they must also be tangents to circle ( O ) since ( $\angle OAP$ ) and ( $\angle OBP$ ) are right angles (tangent-radius theorem).

• Questions for Further Clarification:

  • Is my construction correct and sufficiently rigorous to prove the theorem?
  • Are there alternative methods or simpler geometric constructions that can also demonstrate this principle effectively?

Answer 1

HINTS:

So I left some blanks intentionally for you to fill up in the exercise. A quadratic equation provides an opportunity to include two geometric solutions... here two tangents. Symbols are slightly different.

Solve

$$ y = m x+c, ~~x^2+y^2=r^2~~$$

Slope angle that BP makes to x-axis= $\alpha$

$$ m= \tan \alpha =\pm\sqrt{\frac{c^2}{r^2}-1 }$$

Note the double sign. One sign reserved for each tangent. Plus for blue, minus for green.

The plus and minus signs are proof regarding existence of two solutions as two tangents.

Answer 2

Your construction is not correct. This is because $O$ does not lie on the circle that you constructed. Simply, consider when $OP = 3r$. In this case, $OM=2r \neq PM = r$.

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The right construction would be to draw a circle centered at $O$ with radius $OP$, then a circle centered at $P$ with radius $OP$. Let the two intersection points be $X$ and $Y$. Now, connect $XY$ (the radical axis) and let $XY \cap OP$ be called $M$. Note that this time, $OM=MP$, due to equilateral triangles.

Now, draw a circle centered around $M$ calling it $\Gamma$, intersecting the original circle at points $A$ and $B$. This time, since $O$ is on $\Gamma$, we have $\angle OAP = \angle OBP = 90^\circ$ and so we have constructed tangents.

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As for another construction, this one is convoluted, but works by finding the inverse of the point $A$ with respect to the original circle, regardless if $A$ is inside or outside the circle.

If we adopt your original point names, consider the intersection of point $OA$ with the original circle as $B$. Now, draw a circle with center $B$ and radius $BO$. Let that intersect with the other circle at the point $C$ (there are two intersections, but choose any one of them). Now construct the circle $\Gamma$ with center $C$ and radius $CB$. Let $\Gamma \cap AC = D$. Now draw another circle $\Xi$ with center $B$ and radius $BD$. Let $BD \cap \Xi = A'$. $A'$ would be the inverse of $A$.

Now for your specific scenario, draw a circle with center $A'$ with radius $A'B$. Let this intersect $AO$ at $E$. Draw two circles with radius $BE$, one with center $B$ and one with center $E$. The line $A'E$ would intersect the original circle at $X$ and $Y$, which would be the contact points of the tangents from $A$ to the original circle.