Generation of a cardioid as the locus of intersection point of two tangents to two intersecting circles

Question

About 8 months ago I came up with this nice feature using GeoGebra but I couldn't prove it, any help would be appreciated

We have two intersecting circles and we drew the moving line passing through one of the two intersection points, then we drew the tangents to the two circles at the ends of its intersection with the two circles and set the red point that represents the point of intersection of the two tangents. What I arrived at is that the traces of the red point will create a cardiac curve whose head is the other point of intersection of the two circles and touches both circles.

Answer 1

EDIT. I completely revised my answer, because I found a simpler approach.

Let the centers of the circles (blue in figure below) be $A$, $B$ and let $D$, $E$ be their intersection points. Construct diameters $EM=a$ and $EN=b$ as in the figure. Line $MN$ is parallel to $AB$ and passes through $D$ (because $AB$ intersects $ED$ at the midpoint of $ED$). Hence point $P$, intersection of the tangents at $M$ and $N$, belongs to the locus described in the answer.

Consider now another point $M'$ on the circle with center $A$ and its companion $N'$ on the circle with center $B$, such that line $M'N'$ passes through $D$. If we set $$ \angle MAM'=\theta $$ then $$ {\theta\over2}=\angle MEM'=\angle MDM'=\angle NDN'=\angle NEN'. $$ Quadrilateral $EM'P'N'$ is then cyclic, because: $$ \angle EM'P'={\pi\over2}-{\theta\over2},\quad \angle P'N'E={\pi\over2}+{\theta\over2} $$ and its circumradius is $$ r={M'N'\over2\sin\phi}, $$ where $\phi=\angle M'EN'=\angle MEN$. Hence: $$ EP'=2r\sin{\pi+\theta\over2}={M'N'\over\sin\phi}\cos{\theta\over2}. $$ But triangles $M'EN'$ and $MEN$ are similar, because they have the same angle at vertex $E$ and $EM'/EM=EN'/EN=\cos{\theta\over2}$, hence: $$ M'N'=MN\cos{\theta\over2}. $$ We have then: $$ EP'={MN\over\sin\phi}\cos^2{\theta\over2} ={MN\over2\sin\phi}(1+\cos\theta), $$ which is just the polar equation of a cardioid, if we take ray $EP$ as reference axis. In fact we have $\angle ′=$ because both tangents $′′$ and $′′$ are rotated by $$ with respect to $$ and $$.

Hence the locus of $P'$ is a cardioid, as it was to be proved. Note that $MN/(2\sin\phi)$ is the circumradius of triangle $EMN$, that is twice the circumradius of triangle $EAB$.

In the light of the last remark above, we can check that $P'$ is also the locus traced by a point on the perimeter of a circle, that is rolling around a fixed circle of the same radius (see the definition of cardioid). The fixed circle is the circumcircle of $AEB$, with center $O$ and radius $OE$. The rolling circle (red in figure below) starts with a diameter aligned with $EO$, ending at $P$.

If the red circle is rolled by an angle $\theta$, then point $P$ is carried to $P'$ and it is easy to prove that $\angle PEP'=\theta$ (because $EOCP'$ is a trapezoid) and $$EP'=2OE(1+\cos\theta),$$ the same result we got above.

Answer 2

In M2 using $C_1: p^2+(q-1)^2=3^2$ and $C_2: (p-1)^2+q^2=4^2$ not choosing which intersection point $(p,q)$ to use and with $(h_1,k_1),(h_2,k_2)$ the tangent points for $C_1,C_2$ resp. with $(p,q), (h_1,k_1),(h_2,k_2)$ on the line $\begin{vmatrix}p&q&1\\h_1&k_1&1\\h_2&k_2&1\end{vmatrix}=0$ and finally the two tangents at $(h_1,k_1),(h_2,k_2):$ $$2h_1(x-h_1)+2(k_1-1)(y-k_1)=0,\\2(h_2-1)(x-h_2)+2k_2(y-k_2)=0$$ defining the point $(x,y)$ we want:

R=QQ[h1,k1,h2,k2,p,q,x,y,MonomialOrder => Eliminate 6]
I=ideal(p^2+(q-1)^2-3^2,(p-1)^2+q^2-4^2,h1^2+(k1-1)^2-3^2,(h2-1)^2+k2^2-4^2,determinant matrix({{p,q,1},{h1,k1,1},{h2,k2,1}}),2*h1*(x-h1)+2*(k1-1)*(y-k1),2*(h2-1)*(x-h2)+2*k2*(y-k2))
gens gb I;
toString factor oo_0_0

we get a combined equation over ${\Bbb Q}$ of degree $8=2\cdot 4$ for both cardioids defined much like the two line pairs of degree $2\cdot 1$ of tangents to the circles about the intersection points between the circles (in that to factor further, we step out of ${\Bbb Q}$).

Answer 3

My answer has different levels.

First idea : A reformulation of the question under a generalized form.

Fot this, we need to introduce the lemniscate with polar/implicit equations :

$$r=1-\cos \theta \ \iff \ (x^2+y^2+x)^2=x^2+y^2 \tag{1}$$

right at the beginning ; then we need to consider/remember one of the numerous modes of generation of the cardioid as the envelope of circles $(C_a)$ (see left figure) passing through its apex with polar/implicit equations

$$(C_a): \ \ r=\cos(\theta-a)-\cos(a) \ \iff \ x^2+y^2+(1-\cos a)x-(\sin a)y=0$$

In this perspective, I propose a generalization of the initial issue :

Modified objective : Show that, if one takes any point $L$ on the cardioid, and any pair of circles $(C_a),(C_{a'})$, the external tangents issued from $L$ to these circles are such that their points of tangency are aligned with the point of intersection $I$ of $(C_a)$ and $(C_{a'})$ different from the origin (materialized by a little red circle).

Second idea : use inversion transform in order to get a more tractable problem. Have a look at the correspondence between the left and the right figure, where the colors of the corresponding elements have been preserved (the circle of inversion, i.e., the set of invariant points, is featured in green).

Little recall :

• inversion exchanges lines (not passing through the origin) $\leftrightarrow$ circles passing through the origin.

• inversion preserves the absolute value of angles, therefore preserves in particular tangencies of circles and lines.

This is why the family of circles (internaly) tangent to the cardioid becomes a family of tangent lines "enveloping" the image of the cardioid. This image is the curve with polar/implicit equation :

$$r=\frac{1}{1-\cos \theta} \ \iff \ y^2 = 2x+1$$

where we recognize a parabola with the origin as its focus and parameter $1$.

The blue initial tangent lines become the blue circles passing through the origin and tangent to the red lines. Therefore, the issue boils down to prove that the second intersection point of the two blue circles belongs to the parabola.

Fig. 1 : (reminds the rings of Olympic Games...) LHS: the initial (generalized) situation. RHS : the image of the LHS figure by the inversion transform (inversion circle in green).

Disclaimer : It remains to solve the transformed problem (I haven't had the time to do it).

Remarks :

1. Circles $(C_a)$ have their centers on the circle centered in $(0,-\tfrac12)$ with radius $\tfrac12$.

2. Point $I$ (little red circle), intersection of $(C_a)$ and $(C_{a'})$, can be shown to have an interesting polar angle : $\tfrac12(a+a')$.