# Constructing a circle through a point in the interior of an angle

Is it possible, given a point in the interior of an angle to construct a circle through that point tangent to the rays forming the angle?

Suppose you have a point $A$ in the interior of an angle with vertex $O$. You can bisect the angle, take a point $I$ on the angle bisector, and drop perpendiculars $IP$ and $IQ$ to each side of the angle. The circle with center $I$ and radius $IP$ is then tangent to the rays of the angle. Call this circle $\Gamma$.

You can then draw a line $OA$, which will intersect $\Gamma$ at some point $A'$ on the far side of $\Gamma$ from $O$. You can then draw lines $EA$ parallel to $PA'$ and $FA$ parallel to $QA'$, where $E$ is the intersection with the ray containing $P$, and $F$ is the intersection with the ray containing $Q$. It seems intuitively obvious that the circle through $EAF$ will be a circle through $A$ tangent to the rays of the angle. But is there a way to prove it? I don't see how to do so. Thanks.

## 3 Answers

Let $\Phi$ be the homothety centered at $O$ mapping $A'$ to $A$. Then $\Phi$ fixes the lines $\overline{OP}$ and $\overline{OQ}$, so it maps $P$ to $E$ and $Q$ to $F$. Then it maps the circle $\Gamma$ to the circle through $EAF$, so the circle through $EAF$ must be tangent to $\overline{OP}$ and $\overline{OQ}$

Consider $\triangle OAE$ and $\triangle OA'P$. Because they share vertex $O$ and $AE\parallel A'P$, the triangles are similar. Similar triangles are dilation images of one another, so $\triangle OAE$ is the image of $\triangle OA'P$ under the dilation with center $O$ and ratio $\frac{OA}{OA'}$—call this dilation $T_d$. Likewise, $\triangle OAF$ is the image of $\triangle OA'Q$ under $T_d$. That is, $E$, $A$, and $F$ are the images of $P$, $A'$, and $Q$, respectively, under $T_d$.

Since three points uniquely determine a circle, the circle through $E$, $A$, and $F$ is the image of the circle you called $\Gamma$ under $T_d$, so its center—call it $X$—is the image of $I$ under $T_d$. Now, $T_d$ maps $\triangle OIP$ to $\triangle OXE$, so $\overline{XE}\perp\overline{OE}$, and similarly $\overline{XF}\perp\overline{OF}$, so the circle through $E$, $A$, and $F$ is tangent to the rays of the angle.

This isn't exactly a construction, but this might prove that such a circle exists. The basic idea is to slide a point $p$ on the angle bisector until the length of the dropped perpendicular is equal to the distance between $p$ and $A$. Then $p$ will be the center of a circle tangent to the two rays (and the circle will include $A$ by design). Here goes:

We can assume one of the rays determining the angle is the positive x-axis. So O is the origin, and let $x_0$ be the x-coordinate of your point A. We want to find a point $p=(x_1, y_1)$ on the angle bisector such that $d(A,p)=x_1$.

So let $p=(x_1,y_1)$ be an arbitrary point on the bisector. We have $x_1=0$ when $p=O$ (in particular $x_1<d(A,p)$ ), and we have $x_1>d(A,p)$ when $x_0=x_1$. Since these distances are varying continuously for $x_1\in [0, x_0]$, the IVT implies there is a point $p$ on the bisector with $x_1=d(A,p)$. This is the point you want.

NOTE: I would post a nice picture, but my reputation won't allow it.