
# 15.8: Construction of a polar


In this section we describe a powerful trick that can be used in the constructions with ruler.

Assume $$\Gamma$$ is a circle in the plane and $$P\notin \Gamma$$.Draw two lines $$x$$ and $$y$$ thru $$P$$ that intersect $$\Gamma$$ at two pairs of points $$X$$, $$X'$$ and $$Y$$, $$Y'$$. Let $$Z=(XY) \cap (X'Y')$$ and $$Z'=(XY') \cap(X'Y)$$. Consider the line $$p=(ZZ')$$.

Claim $$\PageIndex{1}$$

The constructed line $$p=(ZZ')$$ does not depend on the choice of the lines $$x$$ and $$y$$.

Moreover, $$P \leftrightarrow p$$ can be extended to a duality such that any point $$P$$ on the circle $$\Gamma$$ corresponds to a line $$p$$ tangent to $$\Gamma$$ at $$P$$.

We will use this claim without a proof, but the proof is not hard. If $$P$$ lies outside of $$\Gamma$$, it can be done by moving $$P$$ to infinity keeping $$\Gamma$$ fixed as a set. If $$P$$ lies inside of $$\Gamma$$, it can be done by moving $$P$$ to the center of $$\Gamma$$. The existence of corresponding projective transformations follow from the idea in Exercise 16.3.1.

The line $$p$$ is called the polar of the point $$P$$ with respect to $$\Gamma$$.

The point $$P$$ is called the pole of the line $$p$$ with respect to $$\Gamma$$.

Exercise $$\PageIndex{1}$$

Revert the described construction. That is, given a circle $$\Gamma$$ and a line $$p$$ that is not tangent to $$\Gamma$$, construct a point $$P$$ such that the described construction for $$P$$ and $$\Gamma$$ produces the line $$p$$.

Hint

Suppose $$p = (QR)$$; denote by $$q$$ and $$r$$ the dual lines produced by the construction. Then, by Claim $$\PageIndex{1}$$, $$P$$ is the point of intersection of $$q$$ and $$r$$.

Exercise $$\PageIndex{2}$$

Let $$p$$ be the polar line of point $$P$$ with respect to the circle $$\Gamma$$. Assume that $$p$$ intersects $$\Gamma$$ at points $$V$$ and $$W$$. Show that the lines $$(PV)$$ and $$(PW)$$ are tangent to $$\Gamma$$.

Come up with a ruler-only construction of the tangent lines to the given circle $$\Gamma$$ thru the given point $$P\not\in \Gamma$$.

Hint

The line $$v$$ polar to $$V$$ is tangent to $$\Gamma$$. Since $$V \in p$$, by Claim $$\PageIndex{1}$$, we get that $$P \in v$$; that is, $$(PV) = v$$. Hence the statement follows.

Exercise $$\PageIndex{3}$$

Assume two concentric circles $$\Gamma$$ and $$\Gamma'$$ are given. Construct the common center of $$\Gamma$$ and $$\Gamma'$$ with a ruler only.

Hint

Choose a point $$P$$ outside of the bigger circle. Construct the lines dual to $$P$$ for both circles. Note that these two lines are parallel.

Assume that the lines intersect the bigger circle at two pairs of points $$X, X'$$ and $$Y, Y'$$. Set $$Z = (XY) \cap (X'Y')$$. Note that the line $$(PZ)$$ passes thru the common center.

The center is the intersection of $$(PZ)$$ and another line constructed the same way.

Exercise $$\PageIndex{4}$$

Assume a line $$\ell$$ and a circle $$\Gamma$$ with its center $$O$$ are given. Suppose $$O\notin \ell$$. Construct a perpendicular from $$O$$ on $$\ell$$ with a ruler only.

Hint

Construct polar liens to two point on $$\ell$$. Denote by $$L$$ the intersection of these two liens. Note that $$\ell$$ is polar to $$L$$ and therefore $$(OL) \perp \ell$$.