0
$\begingroup$

I've tried to prove that there exists only one point $P$ on $O_1O_2$ such that $Pow(P,O_1)=Pow (P,O_2)$ where $O_1 $ and $O_2$ are circles with no point of tangency and I've got the following contradiction:

(Let $R$ and $r$ be the radii of circles $O_1$, and $O_2$ respectively)

$Pow(P,O_1)=Pow(P,O_2)$

$(PO_1)^2- R^2=(PO_2)^2 -r^2$

Then using the fact that $PO_1+PO_2=O_1O_2$ for $PO_1$ , I have:

$(O_1O_2)^2-2PO_2 \cdot O_1O_2 = R^2-r^2$

$O_1O_2(PO_1-PO_2)=R^2-r^2$

Now if I let $PO_1=R+x$ and $PO_2=r+y$ and consequently $O_1O_2=R+r+x+y$ I have

$[(R+x)+(r+y)] \cdot [R+x -(r+y)]=R^2-r^2$

and finally $(R+x)^2-(r+y)^2=R^2-r^2$ which must have solutions $x,y=0$ .

But doesn't that imply $P$ being the point of intersection of circles $O_1,O_2$?What am i doing wrong ?

$\endgroup$
6
  • $\begingroup$ Let's clarify this: $P$ is on the line connecting the centres of $O_1, O_2$? $\endgroup$
    – Element118
    Commented Nov 9, 2015 at 10:53
  • $\begingroup$ Yes ,I forgot to clarify that... $\endgroup$
    – Nameless
    Commented Nov 9, 2015 at 10:53
  • $\begingroup$ So how can $P$ be on the intersection of $O_1,O_2$ if it is on the line connecting the centres? $\endgroup$
    – Element118
    Commented Nov 9, 2015 at 10:55
  • 1
    $\begingroup$ You cannot conclude x and y are both zero from the difference of square equality. $\endgroup$
    – cr001
    Commented Nov 9, 2015 at 10:57
  • $\begingroup$ @ Element118 I don't understand what you say,because that's what i am asking or not ? $\endgroup$
    – Nameless
    Commented Nov 9, 2015 at 10:57

1 Answer 1

1
$\begingroup$

What am i doing wrong ?

I think that we cannot say that $$(R+x)^2-(r+y)^2=R^2-r^2\Rightarrow x=y=0.$$


Let $O_1O_2=s,O_1P=X$. Then, since $PO_2=s-X$, we have $$\begin{align}&O_1O_2(O_1O_2-2PO_2)=R^2-r^2\\&\Rightarrow s(s-2(s-X))=R^2-r^2\\&\Rightarrow s^2-2s(s-X)=R^2-r^2\\&\Rightarrow 2sX=R^2-r^2+s^2\\&\Rightarrow X=\frac{R^2-r^2+s^2}{2s}\end{align}$$ So, there exists only one such point.

$\endgroup$
2
  • $\begingroup$ So i had only to rewrite $y$ in terms of $x$ to get a solution (i've deleted the last comment since it wasn't really usefull). $\endgroup$
    – Nameless
    Commented Nov 9, 2015 at 11:29
  • 1
    $\begingroup$ @Nameless: You can have $(x,y)=(X-R,s-r-X)$. $\endgroup$
    – mathlove
    Commented Nov 9, 2015 at 11:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .