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BERKELEY MATH CIRCLE, October 18 1998

INVERSION IN THE PLANE. PART II: RADICAL AXES¹

Zvezdelina Stankova, UC Berkeley

Definition 2. The degree of point $A$ with respect to a circle $k(O,R)$ is defined as

$$d_k(A)=OA^2-R^2.$$

This is simply the square of the tangent segment from $A$ to $k$. Let $M$ be the midpoint of $AB$ in $\triangle ABC$, and $CH$ - the altitude from $C$, with $H\in AB$ (cf. Fig.5-6.) Mark the sides $BC$, $CA$ and $AB$ by $a$, $b$ and $c$, respectively. Then

$$\vert a^2-b^2\vert=\vert BH^2-AH^2\vert=c\vert BH-AH\vert=2c\cdot MH,$$ (1)

where $M$ is the midpoint of $AB$.

Definition 3. The radical axis of two circles $k_1$ and $k_2$ is the geometric place of all points which have the same degree with respect to $k_1$ and $k_2$: $\{A\,\vert\,d_{k_1}(A)=d_{k_2}(A)\}$.

Let $P$ be one of the points on the radical axis of $k_1(O_1,R_1)$ and $k_2(O_2,R_2)$ (cf. Fig.7.) We have by ():

$$PO_1^2-R_1^2=PO_2^2-R_2^2\,\,\Rightarrow\,\,\vert R_1^2-R_2^2\vert=\vert PO_1^2-PO_2^2\vert =2O_1O_2\cdot MH,$$

where $M$ is the midpoint of $O_1O_2$, and $H$ is the orthogonal projection of $P$ onto $O_1O_2$. Then

$$MH=\frac{\vert R_1^2-R_2^2\vert}{2O_1O_2}=$$constant$$\,\, \Rightarrow\,\,$$point$$\,\,H\,\,$$is constant$$.$$

(Show that the direction of $MH^{\rightarrow}$ is the same regardless of which point $P$ on the radical axis we have chosen.) Thus, the radical axis is a subset of a line $\perp O_1O_2$. The converse is easy.

Lemma 1   Let $k_1(O_1,R_1)$ and $k_2(O_2,R_2)$ be two nonconcentric circles circles, with $R_1\geq R_2$, and let $M$ be the midpoint of $O_1O_2$. Let $H$ lie on the segment $MO_2$, so that

$$HM=(R_1^2-R_2^2)/2O_1O_2.$$

Then the radical axis of $k_1(O_1,R_1)$ and $k_2(O_2,R_2)$ is the line $l$, perpendicular to $O_1O_2$ and passing through $H$.

What happens with the radical axis when the circles are concentric? In some situations it is convenient to have the circles concentric. In the following fundamental lemma, we achieve this by applying both ideas of inversion and radical axis.

Lemma 2   Let $k_1$ and $k_2$ be two nonintersecting circles. Prove that there exists an inversion sending the two circles into concentric ones.

PROOF: If the radical axis intersects $O_1O_2$ in point $H$, let $k(H,$d$_{k_i}(H))$ intersect $O_1O_2$ in $A$ and $B$. Apply inversion wrt $k^{\prime}(A,AB)$ (cf. Fig. 8.) Then $I(k)$ is a line $l$ through $B$, $l\perp O_1O_2$. But $k_1\perp k$, hence $I(k_1)\perp l$, i.e. the center of $I(k_1)$ lies on $l$. It also lies on $O_1O_2$, hence $I(k_1)$ is centered at $B$. Similarly, $I(k_2)$ is centered at $B$. $\qedsymbol$

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