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Two circles $c$ and $d$ that intersect at points $A$ and $B$ are given. Let $p$ be a line passing through $A$ that intersects circles $c$ and $d$ at points $P_1$ and $P_2$. Construct a circle tangential to $p$ and orthogonal to both $c$ and $d$.

enter image description here

What I did was that I constructed a circle $k=k(P, |PS_1|)$ so that $P$ is the midpoint of $S_1S_2$. Then I applied inversion about $k$, which maps circles $c$ and $d$ to two parallel lines $c'$ and $d'$, and line $p$ to circle $p'$ that passes through point $P$. Then I found line $l$ such that it's perpendicular to $c'$ and $d'$ and tangential to circle $p '$. Finally, I applied inversion about $k$ again so it would map $l$ to circle $l'$. I thought $l'$ would be the solution, but apparently it isn't. Why doesn't this work and what should I do instead?

EDIT: Now I see what a dumb mistake I made, so I'm sorry for hurting yalls eyes. I still wonder how this problem could be solved using inversion. I was thinking to apply inversion about circle centered at $A$ that would map both $c$ and $d$ to lines $c'$ and $d'$ and it would map line $p$ to itself, but I don't see how that would be helpful

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3 Answers 3

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In fact, after having presented an analytical solution, I have discovered a "pure geometry" simple construction using inversion.

enter image description here

Let us take the notations of figure below $O_1,O_2$ for the centers of circles $(C_1),(C_2)$ resp. with intersection points $A,A'$, $AD_1,AD_2$ for diameters.

Let us proceed by necessary condition first. Let $F$ be the center of orthogonal circle $(C)$ to both circles $(C_1),(C_2)$ with radius $FE \perp (p)$. Let $(I)$ be the inversion with inversion circle $(C)$. The two circles $(C_1),(C_2)$, being orthogonal to $(C)$ are globally invariant by $(I)$. Therefore $A,A'$ are exchanged by $(I)$, which means two properties : $A,A',F$ are aligned and

$$FA.FA'=FE^2\tag{1}$$

But (1) is possible iff $EB$ is the altitude issued from $E$ in the right triangle $AEF$ (see second "mean proportional" property here) which is a characteristic property of the main altitude of a right triangle.

Whence the construction of point $F$ :

Draw the perpendicular line in $A'$ to $AA'$ (aka radical axis of circles $(C_1),(C_2)$). This line intersects line (p) in $E$. Then erect in $E$ the perpendicular line to (p). This line intersects line $AA'$ in $F$, and it's finished.

Once $F$ is known, the radius of the orthogonal circle is of course $FE$.

Remark : I still don't understand the construction of the other answer.

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  • $\begingroup$ Any comment ? ... $\endgroup$
    – Jean Marie
    Commented Aug 14, 2023 at 6:24
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    $\begingroup$ You are not the only one who cannot fathom the accepted answer. Check this and this accepted answer by the same answerer and see whether you can understand them. Probably you will see some kind of pattern of answering. Reputation of MSE is at stake. $\endgroup$
    – YNK
    Commented Aug 20, 2023 at 17:01
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enter image description here

Hint: the center of such a circle is on the radical axis EF of two circles. Line l intersect the line connecting the centers A and C of two circle at Q. Draw bisector of angle PQA, it intersect radical axis at G. This is the center of circle. Draw two tangents on two circles, the circle center at G passing the tangent points will be orthogonal on two circle ang tangent on line l. Could you explain the reason?

Here is figures for other case:

enter image description here

enter image description here

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    $\begingroup$ @Katarina, You are welcome. Well understood the reason., +1. $\endgroup$
    – sirous
    Commented Aug 4, 2023 at 7:56
  • $\begingroup$ Sorry but I don't understand your construction with bissector line. Please explain it. It would mean that the looked-for orthogonal circle is tangent to line AC which is not the case... $\endgroup$
    – Jean Marie
    Commented Aug 9, 2023 at 21:26
  • $\begingroup$ @sirous kindly, please answer the question in my previous comment. $\endgroup$
    – Jean Marie
    Commented Aug 11, 2023 at 6:06
  • $\begingroup$ @JeanMarie, Sorry for delay, I was in vacation. The circle is tangent to bothe AC and ex tension of EF as required. $\endgroup$
    – sirous
    Commented Aug 20, 2023 at 17:23
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I am aware that this question has been asked in a pure geometry context.

Nevertheless, an analytic approach has its own merit.

This is what I would like to show here.


Preliminaries :

Consider the two forms of the equation of a circle with center $(a,b)$ and radius $R$:

$$\begin{cases} x^2+y^2-2ax-2by+c&=&0\\ (x-a)^2+(y-b)^2&=&R^2 \end{cases}$$

Their comparison brings formula :

$$a^2+b^2-c=R^2\tag{1}$$

We will call $[a,b,c]$ the "characteristics" of the corresponding circle.

Remark : A particular case : due to formula (1), a circle passing through the origin has its characteristics under the form $[a,b,0]$.

Let us recall a classical orthogonality criteria (see formula (4) here) for 2 circles with characteristics $[a_0,b_0,c_0]$ and $[a_1,b_1,c_1]$ :

$$a_0a_1+b_0b_1-\frac12(c_0+c_1)=0 \tag{2}$$

(end of preliminaries)


enter image description here

Let us now consider a cartesian coordinate system such that point $A$ is the origin and line (d) is the axis of abscissas (see figure below).

The characteristics of these circles are therefore $[a_k,b_k,\color{red}{0}], \ \ k=1,2$, (see Remark above).

Let $(a_0,b_0)$ be the center of the looked-for orthogonal circle to both of them.

This orthogonality is expressed through formula (2) and the same formula with indices "1", replaced by indices "2", giving :

$$\begin{cases}a_0a_1+b_0b_1-\frac12(c_0+0)&=&0\\ a_0a_2+b_0b_2-\frac12(c_0+0)&=&0\end{cases}$$

which is a system of 2 linear equations with 3 unknowns $a_0,b_0,c_0$ with a one-dimensional set of solutions :

$$\begin{cases}a_0&=& \ \ \ k(b_2-b_1)\\b_0&=&-k(a_2-a_1)\\c_0&=&2k(a_1b_2-a_2b_1)\end{cases}\tag{3}$$

for a certain $k$ that we are going to find now.

But before that, let us consider the first two equations in (3) : they express that the center $(a_0,b_0)$ belongs to the line issued from the origin, orthogonal to the line joining the centers ; otherwise said, the center $(a_0,b_0)$ is on the radical axis of the two circles.

It remains to express that this circle with characteristics $[a_0,b_0,c_0]$ is tangent to line (d) which is (let us recall it) the abscissas' axis. This is easy : its radius must be $R_0=|b_0|$, a fact that will give the following relationship, due to formula (1) :

$$a_0^2+b_0^2-c_0=b_0^2 \ \iff \ a_0^2=c_0$$

giving, thanks to (3):

$$k^2(b_2-b_1)^2=2k(a_1b_2-a_2b_1) \ \iff \ k=2 \frac{a_1b_2-a_2b_1}{(b_2-b_1)^2} \tag{4}$$

Now, this orthogonal circle is completely determined : more precisely, it exists (and is unique) under the condition (see the denominator in (4)) that $b_1 \ne b_2$, i.e., the line joining the centers must not be parallel to line (d).

The graphical representation has been obtained with the Matlab program below :

clear all;close all;hold on;axis equal off;
plot([-5,20],[0,0],'k');
plot([0,0],[-10,20],'k');
a1=2;b1=4;k1=a1+i*b1; % center n°1
a2=7;b2=-2;k2=a2+i*b2 % center n°2
plot([k1,k2],'-ob')
u=exp(i*(0:0.01:2*pi)); % unit circle
plot(k1+abs(k1)*u,'r'); % first circle (red)
plot(k2+abs(k2)*u,'r'); % second circle (red)
m=2*(a1*b2-a2*b1)/(b2-b1)^2;
a0=m*(b2-b1);b0=-m*(a2-a1);k0=a0+i*b0;
plot(k0+b0*u,'k') ; % orthogonal circle (black)
plot([0+0i,k0],'-ok');
text([a0,a1,a2]+[1,-4,1],[b0,b1,b2]+[1,0,0],...
{'(a_0,b_0)','(a_1,b_1)','(a_2,b_2)'})
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