Given a point $P$ and two circles $\Gamma$ and $\Gamma'$ with centers $O$ and $O'$. I'm searching for a method to construct lines passing through $P$ and defining on each circle chords of same length.
A geometer professor made the following drawing to illustrate the problem.
Many thanks for any constructive idea or suggestions.
On
Suppose first that the two circles intersect each other. Let $F$ be the midpoint of the line of centres $O_1O_2$, and let $F'$ be the reflection of $F$ in the radical axis $\mathcal{R}$ of the two circles. Let $\mathcal{P}$ be the parabola with focus $F$ and directrix the line through $F'$ parallel to the radical axis $\mathcal{R}$. If $\ell$ is a line, let $X_\ell$ be the foot of the perpendicular from $F$ to $\ell$. Then $\ell$ is a tangent to $\mathcal{P}$ if and only if $X_\ell$ lies on the radical axis.
Given a point $P$, suppose that $\ell$ is a tangent to $\mathcal{P}$ that passes through $P$. Let $M_1$, $M_2$ be the feet of the perpendiculars from $O_1$, $O_2$ to the line $\ell$. The lines $O_1M_1$, $FX_\ell$, $O_2M_2$ are parallel, and $F$ is the midpoint of $O_1O_2$, and hence $X_\ell$ is the midpoint of $M_1M_2$. If this line $\ell$ intersects the two circles, forming chords of lengths $w_1$, $w_2$ respectively (and midpoints $M_1$, $M_2$), then the Intersecting Chords Theorem tells us that
$$ X_\ell M_1^2 - \frac14w_1^2 \; = \; X_\ell A \times X_\ell B \; = \; X_\ell M_2^2 - \frac14w_2^2 $$
where the radical axis meets the two circles at $A$ and $B$, and hence $w_1=w_2$.
Start with a point $P$ outside the parabola $\mathcal{P}$, and draw a circle with $FP$ as diameter. Since $P$ lies outside the parabola $\mathcal{P}$, this circle will intersect the radical axis $\mathcal{R}$ at two points $X_1$, $X_2$. The lines $PX_1$, $PX_2$ will be tangents to $\mathcal{P}$; if the lines intersect both circles, they will cut the circles in congruent chords with midpoints $M_{1,1},M_{2,1}$ and $M_{1,2},M_{2,2}$ respectively.
It is easy to show that the mutual tangents to the two circles, which pass through the external similitude centre $Se$ of the circles, are themselves tangents to $\mathcal{P}$. It is clear that the line $PX_1$ will not intersect the two circles if $P$ lies above the upper mutual tangent, and that $PX_2$ will not intersect the two circles if $P$ lies below the lower mutual tangent.
For us to be able to perform this construction, therefore, $P$ must lie outside the parabola $\mathcal{P}$ (so that the points $X_1,X_2$ exist) and also $P$ must lie outside the wedge formed by the two mutual tangents to the circles vertically opposite to $\mathcal{P}$.
If the two circles do not intersect, the basic construction remains the same, but the allowed region for $P$ is further limited. Consideration also needs to be made of where $P$ lies in relation two the two mutual tangents to the circle that meet at their internal similitude centre.
Let $r_1$ and $R_1$ the radii of $\Gamma$ and $\Gamma '$ respectively, and $w$ the length of the desired chord.
Construct two circles $\Lambda$ and $\Lambda '$ concentric to $\Gamma$ and $\Gamma '$ with radii $r$ and $R$ respectively, so that: $$r = \sqrt{r_1^2-(\frac{w}{2})^2} \quad \quad (1)$$ and $$R = \sqrt{R_1^2-(\frac{w}{2})^2}\quad \quad (2)$$ Let $OP=d$, $O'P=D$,and $OO'=L$.
Let $PS$ the height of $\triangle POO'$ relative to side $OO'$, and let's call $PS=H$ and $O'S=\delta$ for short. We have:
$$\delta = \frac{D^2-d^2+L^2}{2L} \quad \quad (3)$$
and
$$H^2 = D^2 - \delta^2 \quad \quad (4)$$
The value of $w$, so that P is a point of the secant lines that does not cross the segment $OO'$, can be found solving equations (1), (2), (3), (4) e (5).
$$\frac{LR}{R-r} -\delta = \sqrt{(\frac{HL} {R-r})^2-R^2} \quad \quad (5)$$
That value of $w$ makes the two secant lines relative to $\Gamma$ and $\Gamma '$ be also tangent lines of $\Lambda$ and $\Lambda '$.