# Euclidean Geometry Intersection of Circles

Two circles intersect in the Cartesian Coordinate system at points $A$ and $B$. Point $A$ lies on the line $y=3$. Point $B$ lies on the line $y=12$. These two circles are also tangent to the x-axis at points $P$ and $Q$. How would one find the distance of $AB$ in terms of $PQ$? I managed to prove that $\angle PAQ+\angle PBQ=180$ (edited from before thanks to J.D) but i dont know if that gets anywhere. Thanks!

-
Are you sure that $\angle PAQ=\angle PBQ$? I might be wrong but here is how I picture this problem in my mind drawing. –  user2468 Feb 25 '12 at 21:08
Im sorry, i mean to say $\angle PAQ + \angle PBQ=180$, so the circles around $PAQ$ and around $PBC$ are congruent. I have edited my post –  Abby Feb 25 '12 at 21:20

The symmetries will be clearer if we work more generally, instead of using the specific numbers $3$ and $12$. Suppose also that $PQ=2e$. (We use $2e$ to avoid fractions later.)

Let our two circles meet at points $A$ and $B$, where $A$ is on the line $y=a$ and $B$ is on the line $y=b$. Assume that $a$ and $b$ are positive. We want to find the length of $AB$ in terms of $a$, $b$, and $2e$.

We use analytic geometry. It will turn out that the analytic geometry points to a synthetic argument, but we will not write out the synthetic version.

We can choose coordinates (or slide the circles) so that $P=(-e,0)$ and $Q=(e,0)$. Then one circle has centre $(-e,r)$, where $r$ is the radius of that circle. If the radius of the second circle is $R$, then the second circle has centre $(e, R)$. The equations of the two circles are $$(x+e)^2+(y-r)^2 =r^2 \qquad\text{and}\qquad (x-e)^2+(y-R)^2=R^2.$$

Let $A=(s,a)$ and let $B=(t,b)$. If we knew $s$ and $t$, we would know everything. Because our two points lie on the first circle,
$$(s+e)^2+(a-r)^2=r^2\qquad\text{and}\qquad (t+e)^2+(b-r)^2=r^2. \qquad (1)$$ Because our two points lie on the second circle,
$$(s-e)^2+(a-R)^2=R^2\qquad\text{and}\qquad (t-e)^2+(b-R)^2=R^2. \qquad (2)$$
We manipulate Equations $(1)$ a little, by expanding the terms $(a-r)^2$ and $(b-r)^2$. Quickly we arrive at $(s+e)^2+a^2=2ar$ and $(t+e)^2+b^2=2br$. If we multiply both sides of the first equation by $a$, and both sides of the second equation by $b$, we find that $$b[(s+e)^2+a^2]=a[t+e)^2+b^2].\qquad (3)$$ Similar manipulations on the equations in $(2)$ give $$b[(s-e)^2+a^2]=a[t-e)^2+b^2].\qquad (4)$$ Look now at equations $(3)$ and $(4)$, and subtract. There is very nice cancellation. We get $4bse=4ate$, and therefore $bs=at$.

We can therefore write $s=ak$ and $t=bk$ for some $k$. Substitute for $s$ and $t$ in $(3)$. After some manipulation we arrive at $$ab(a-b)k^2=(a-b)e^2 -ab(a-b).$$ Cancel the $a-b$, and divide by $ab$. We get $$k^2=\frac{e^2}{ab}-1.\qquad (5).$$ Note that $(AB)^2=(s-t)^2+(a-b)^2$. Since $s-t=ak-bk$, we have $$(AB)^2=(a-b)^2k^2+(a-b)^2=(a-b)^2(k^2+1)=(a-b)^2\frac{e^2}{ab}.$$ We conclude that $$AB=\frac{|a-b|}{\sqrt{ab}}e.$$ There is a caveat. In $(5)$, we had an expression for $k^2$. This cannot be negative. So the argument breaks down, and the circles cannot obey the stated conditions, if $e^2<ab$.

Remark: Now that we have seen the answer, and how simple it is, and how $s$ and $t$ are in the proportion $a:b$, we can put together a quicker synthetic proof. Since, for algebraic reasons, it was convenient to let $P$ and $Q$ be at $(-e,0)$ and $(e,0)$, the midpoint of $PQ$ must have geometric significance. The details of a geometric argument are left to you!

There is a very nice theorem due to Tarski that says that there is an algorithm for determining the truth of sentences of elementary geometry. The algorithm works through coordinatization. So it is no accident that there is an analytic geometry solution. That solution can in principle be arrived at mechanically. For the sake of making this task human-feasible, we tried to keep calculations symmetrical.

-
Andre Nicolas, you mentioned in another posting of this question that there was a nicer way to do this problem. Could you please enlighten us on that other method? –  Ali Mar 1 '12 at 21:31

Extend $AB$ so that it hits $PQ$ at $M$. Then by Power of a Point, $MA(MA+AB)=MP^2=MQ^2$ or $MP=MQ$, so we will say that $MP=MQ=e$, following the notation of above.

Now we we drop point $B$ to the line $y=b$ and continue that line to the $x-$axis and note the similar right triangle that arises. Thus the ratio of $AB$ to $MA+AB$ is equal to $a-b$ to $a$ (assuming $a>b$) by similar triangles . Again, this is following the notation of above. Solving for $MA$ we find that $MA=\frac{b}{a-b} \cdot AB$ after some routine computation.

Now returning to $MA(MA+AB)=MP^2=e^2$ (recall $2e=PQ$), we substitute to find $(\frac{b}{a-b}\cdot AB )( \frac{b}{a-b } \cdot AB +AB)=e^2$ and from there the trivial computation is left to the reader.

EDIT: Hopefully my proof makes sense. If not, just to clarify, here is the following notation I use:

$M$ proved to be the midpoint of $PQ$

$B$ is on $y=a$ and $A$ is on $y=b$ with $a>b$

Hopefully the reader can piece together the beautiful solution through my murky display of it. It really is a gem.

-
M has been proved to be the midpoint of PQ in your second sentence and hence need not be iterated again. –  Mick Nov 11 '13 at 8:24