Proof for the parametric representation

A proof can be established using complex numbers and their common description as the complex plane. The rolling movement of the black circle on the blue one can be split into two rotations. In the complex plane a rotation around point ${\displaystyle 0}$ (the origin) by an angle ${\displaystyle \varphi }$ can be performed by multiplying a point ${\displaystyle z}$ (complex number) by ${\displaystyle e^{i\varphi }}$. Hence

the rotation ${\displaystyle \Phi _{+}}$ around point ${\displaystyle a}$ is${\displaystyle :z\mapsto a+(z-a)e^{i\varphi }}$,

the rotation ${\displaystyle \Phi _{-}}$ around point ${\displaystyle -a}$ is: ${\displaystyle z\mapsto -a+(z+a)e^{i\varphi }}$.

A point ${\displaystyle p(\varphi )}$ of the cardioid is generated by rotating the origin around point ${\displaystyle a}$ and subsequently rotating around ${\displaystyle -a}$ by the same angle ${\displaystyle \varphi }$:

$${\displaystyle p(\varphi )=\Phi _{-}(\Phi _{+}(0))=\Phi _{-}\left(a-ae^{i\varphi }\right)=-a+\left(a-ae^{i\varphi }+a\right)e^{i\varphi }=a\;\left(-e^{i2\varphi }+2e^{i\varphi }-1\right).}$$

From here one gets the parametric representation above:

$${\displaystyle {\begin{array}{cclcccc}x(\varphi )&=&a\;(-\cos(2\varphi )+2\cos \varphi -1)&=&2a(1-\cos \varphi )\cdot \cos \varphi &&\\y(\varphi )&=&a\;(-\sin(2\varphi )+2\sin \varphi )&=&2a(1-\cos \varphi )\cdot \sin \varphi &.&\end{array}}}$$

(The trigonometric identities ${\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi ,\ (\cos \varphi )^{2}+(\sin \varphi )^{2}=1,}$ ${\displaystyle \cos(2\varphi )=(\cos \varphi )^{2}-(\sin \varphi )^{2},}$ and ${\displaystyle \sin(2\varphi )=2\sin \varphi \cos \varphi }$ were used.)

Chords through the cusp

C1

Chords through the cusp of the cardioid have the same length ${\displaystyle 4a}$.

C2

The midpoints of the chords through the cusp lie on the perimeter of the fixed generator circle (see picture).

Proof of C1

The points ${\displaystyle P:p(\varphi ),\;Q:p(\varphi +\pi )}$ are on a chord through the cusp (=origin). Hence

$${\displaystyle {\begin{aligned}|PQ|&=r(\varphi )+r(\varphi +\pi )\\&=2a(1-\cos \varphi )+2a(1-\cos(\varphi +\pi ))=\cdots =4a\end{aligned}}.}$$

Proof for C2

For the proof the representation in the complex plane (see above) is used. For the points

$${\displaystyle P:\ p(\varphi )=a\,\left(-e^{i2\varphi }+2e^{i\varphi }-1\right)}$$

and

$${\displaystyle Q:\ p(\varphi +\pi )=a\,\left(-e^{i2(\varphi +\pi )}+2e^{i(\varphi +\pi )}-1\right)=a\,\left(-e^{i2\varphi }-2e^{i\varphi }-1\right),}$$

the midpoint of the chord ${\displaystyle PQ}$ is

$${\displaystyle M:\ {\tfrac {1}{2}}(p(\varphi )+p(\varphi +\pi ))=\cdots =-a-ae^{i2\varphi }}$$

which lies on the perimeter of the circle with midpoint ${\displaystyle -a}$ and radius ${\displaystyle a}$ (see picture).

Cardioid as inverse curve of a parabola

Cardioid generated by the inversion of a parabola across the unit circle (dashed)

A cardioid is the inverse curve of a parabola with its focus at the center of inversion (see graph)

For the example shown in the graph the generator circles have radius ${\textstyle a={\frac {1}{2}}}$. Hence the cardioid has the polar representation

$${\displaystyle r(\varphi )=1-\cos \varphi }$$

and its inverse curve

$${\displaystyle r(\varphi )={\frac {1}{1-\cos \varphi }},}$$

which is a parabola (s. parabola in polar coordinates) with the equation ${\textstyle x={\tfrac {1}{2}}\left(y^{2}-1\right)}$ in Cartesian coordinates.

Remark: Not every inverse curve of a parabola is a cardioid. For example, if a parabola is inverted across a circle whose center lies at the vertex of the parabola, then the result is a cissoid of Diocles.

Cardioid as envelope of a pencil of circles

Cardioid as envelope of a pencil of circles

In the previous section if one inverts additionally the tangents of the parabola one gets a pencil of circles through the center of inversion (origin). A detailed consideration shows: The midpoints of the circles lie on the perimeter of the fixed generator circle. (The generator circle is the inverse curve of the parabola's directrix.)

This property gives rise to the following simple method to draw a cardioid:

1. Choose a circle ${\displaystyle c}$ and a point ${\displaystyle O}$ on its perimeter,
2. draw circles containing ${\displaystyle O}$ with centers on ${\displaystyle c}$, and
3. draw the envelope of these circles.

Proof with envelope condition The envelope of the pencil of implicitly given curves

$${\displaystyle F(x,y,t)=0}$$

with parameter ${\displaystyle t}$ consists of such points ${\displaystyle (x,y)}$ which are solutions of the non-linear system

$${\displaystyle F(x,y,t)=0,\quad F_{t}(x,y,t)=0,}$$

which is the envelope condition. Note that ${\displaystyle F_{t}}$ means the partial derivative for parameter ${\displaystyle t}$.

Let ${\displaystyle c}$ be the circle with midpoint ${\displaystyle (-1,0)}$ and radius ${\displaystyle 1}$. Then ${\displaystyle c}$ has parametric representation ${\displaystyle (-1+\cos t,\sin t)}$. The pencil of circles with centers on ${\displaystyle c}$ containing point ${\displaystyle O=(0,0)}$ can be represented implicitly by

$${\displaystyle F(x,y,t)=(x+1-\cos t)^{2}+(y-\sin t)^{2}-(2-2\cos t)=0,}$$

which is equivalent to

$${\displaystyle F(x,y,t)=x^{2}+y^{2}+2x\;(1-\cos t)-2y\;\sin t=0\;.}$$

The second envelope condition is

$${\displaystyle F_{t}(x,y,t)=2x\;\sin t-2y\;\cos t=0.}$$

One easily checks that the points of the cardioid with the parametric representation

$${\displaystyle x(t)=2(1-\cos t)\cos t,\quad y(t)=2(1-\cos t)\sin t}$$

fulfill the non-linear system above. The parameter ${\displaystyle t}$ is identical to the angle parameter of the cardioid.

Cardioid as envelope of a pencil of lines

Cardioid as envelope of a pencil of lines

A similar and simple method to draw a cardioid uses a pencil of lines. It is due to L. Cremona:

1. Draw a circle, divide its perimeter into equal spaced parts with ${\displaystyle 2N}$ points (s. picture) and number them consecutively.
2. Draw the chords: ${\displaystyle (1,2),(2,4),\dots ,(n,2n),\dots ,(N,2N),(N+1,2),(N+2,4),\dots }$. (That is, the second point is moved by double velocity.)
3. The envelope of these chords is a cardioid.

Cremona's generation of a cardioid

Proof

The following consideration uses trigonometric formulae for ${\displaystyle \cos \alpha +\cos \beta }$, ${\displaystyle \sin \alpha +\sin \beta }$, ${\displaystyle 1+\cos 2\alpha }$, ${\displaystyle \cos 2\alpha }$, and ${\displaystyle \sin 2\alpha }$. In order to keep the calculations simple, the proof is given for the cardioid with polar representation ${\displaystyle r=2(1\mathbin {\color {red}+} \cos \varphi )}$ (§ Cardioids in different positions).

Equation of the tangent of the cardioid with polar representation r = 2(1 + cos 𝜑)

From the parametric representation

$${\displaystyle {\begin{aligned}x(\varphi )&=2(1+\cos \varphi )\cos \varphi ,\\y(\varphi )&=2(1+\cos \varphi )\sin \varphi \end{aligned}}}$$

one gets the normal vector ${\displaystyle {\vec {n}}=\left({\dot {y}},-{\dot {x}}\right)^{\mathsf {T}}}$. The equation of the tangent ${\displaystyle {\dot {y}}(\varphi )\cdot (x-x(\varphi ))-{\dot {x}}(\varphi )\cdot (y-y(\varphi ))=0}$ is:

$${\displaystyle (\cos 2\varphi +\cos \varphi )\cdot x+(\sin 2\varphi +\sin \varphi )\cdot y=2(1+\cos \varphi )^{2}\,.}$$

With help of trigonometric formulae and subsequent division by ${\textstyle \cos {\frac {1}{2}}\varphi }$, the equation of the tangent can be rewritten as:

$${\displaystyle \cos({\tfrac {3}{2}}\varphi )\cdot x+\sin \left({\tfrac {3}{2}}\varphi \right)\cdot y=4\left(\cos {\tfrac {1}{2}}\varphi \right)^{3}\quad 0<\varphi <2\pi ,\ \varphi \neq \pi .}$$

Equation of the chord of the circle with midpoint (1, 0) and radius 3

For the equation of the secant line passing the two points ${\displaystyle (1+3\cos \theta ,3\sin \theta ),\ (1+3\cos {\color {red}2}\theta ,3\sin {\color {red}2}\theta ))}$ one gets:

$${\displaystyle (\sin \theta -\sin 2\theta )x+(\cos 2\theta -\sin \theta )y=-2\cos \theta -\sin(2\theta )\,.}$$

With help of trigonometric formulae and the subsequent division by ${\textstyle \sin {\frac {1}{2}}\theta }$ the equation of the secant line can be rewritten by:

$${\displaystyle \cos \left({\tfrac {3}{2}}\theta \right)\cdot x+\sin \left({\tfrac {3}{2}}\theta \right)\cdot y=4\left(\cos {\tfrac {1}{2}}\theta \right)^{3}\quad 0<\theta <2\pi .}$$

Conclusion

Despite the two angles ${\displaystyle \varphi ,\theta }$ have different meanings (s. picture) one gets for ${\displaystyle \varphi =\theta }$ the same line. Hence any secant line of the circle, defined above, is a tangent of the cardioid, too:

The cardioid is the envelope of the chords of a circle.

Remark:
The proof can be performed with help of the envelope conditions (see previous section) of an implicit pencil of curves:

$${\displaystyle F(x,y,t)=\cos \left({\tfrac {3}{2}}t\right)x+\sin \left({\tfrac {3}{2}}t\right)y-4\left(\cos {\tfrac {1}{2}}t\right)^{3}=0}$$

is the pencil of secant lines of a circle (s. above) and

$${\displaystyle F_{t}(x,y,t)=-{\tfrac {3}{2}}\sin \left({\tfrac {3}{2}}t\right)x+{\tfrac {3}{2}}\cos \left({\tfrac {3}{2}}t\right)y+3\cos \left({\tfrac {1}{2}}t\right)\sin t=0\,.}$$

For fixed parameter t both the equations represent lines. Their intersection point is

$${\displaystyle x(t)=2(1+\cos t)\cos t,\quad y(t)=2(1+\cos t)\sin t,}$$

which is a point of the cardioid with polar equation ${\displaystyle r=2(1+\cos t).}$

Cardioid as caustic: light source ${\displaystyle Z}$, light ray ${\displaystyle {\vec {s}}}$, reflected ray ${\displaystyle {\vec {r}}}$

Cardioid as caustic of a circle with light source (right) on the perimeter

Cardioid as caustic of a circle

The considerations made in the previous section give a proof that the caustic of a circle with light source on the perimeter of the circle is a cardioid.

If in the plane there is a light source at a point ${\displaystyle Z}$ on the perimeter of a circle which is reflecting any ray, then the reflected rays within the circle are tangents of a cardioid.

Proof

As in the previous section the circle may have midpoint ${\displaystyle (1,0)}$ and radius ${\displaystyle 3}$. Its parametric representation is

$${\displaystyle c(\varphi )=(1+3\cos \varphi ,3\sin \varphi )\ .}$$

The tangent at circle point ${\displaystyle C:\ k(\varphi )}$ has normal vector ${\displaystyle {\vec {n}}_{t}=(\cos \varphi ,\sin \varphi )^{\mathsf {T}}}$. Hence the reflected ray has the normal vector ${\displaystyle {\vec {n}}_{r}=\left(\cos {\color {red}{\tfrac {3}{2}}}\varphi ,\sin {\color {red}{\tfrac {3}{2}}}\varphi \right)^{\mathsf {T}}}$ (see graph) and contains point ${\displaystyle C:\ (1+3\cos \varphi ,3\sin \varphi )}$. The reflected ray is part of the line with equation (see previous section)

$${\displaystyle \cos \left({\tfrac {3}{2}}\varphi \right)x+\sin \left({\tfrac {3}{2}}\varphi \right)y=4\left(\cos {\tfrac {1}{2}}\varphi \right)^{3}\,,}$$

which is tangent of the cardioid with polar equation

$${\displaystyle r=2(1+\cos \varphi )}$$

from the previous section.

Remark: For such considerations usually multiple reflections at the circle are neglected.

Cardioid as pedal curve of a circle

Point of cardioid is foot of dropped perpendicular on tangent of circle

The Cremona generation of a cardioid should not be confused with the following generation:

Let be ${\displaystyle k}$ a circle and ${\displaystyle O}$ a point on the perimeter of this circle. The following is true:

The foots of perpendiculars from point ${\displaystyle O}$ on the tangents of circle ${\displaystyle k}$ are points of a cardioid.

Hence a cardioid is a special pedal curve of a circle.

Proof

In a Cartesian coordinate system circle ${\displaystyle k}$ may have midpoint ${\displaystyle (2a,0)}$ and radius ${\displaystyle 2a}$. The tangent at circle point ${\displaystyle (2a+2a\cos \varphi ,2a\sin \varphi )}$ has the equation

$${\displaystyle (x-2a)\cdot \cos \varphi +y\cdot \sin \varphi =2a\,.}$$

The foot of the perpendicular from point ${\displaystyle O}$ on the tangent is point ${\displaystyle (r\cos \varphi ,r\sin \varphi )}$ with the still unknown distance ${\displaystyle r}$ to the origin ${\displaystyle O}$. Inserting the point into the equation of the tangent yields

$${\displaystyle (r\cos \varphi -2a)\cos \varphi +r\sin ^{2}\varphi =2a\quad \rightarrow \quad r=2a(1+\cos \varphi )}$$

which is the polar equation of a cardioid.

Remark: If point ${\displaystyle O}$ is not on the perimeter of the circle ${\displaystyle k}$, one gets a limaçon of Pascal.

Proof

For the cardioid with parametric representation

$${\displaystyle x(\varphi )=2a(1-\cos \varphi )\cos \varphi =4a\sin ^{2}{\tfrac {\varphi }{2}}\cos \varphi \,,}$$

$${\displaystyle y(\varphi )=2a(1-\cos \varphi )\sin \varphi =4a\sin ^{2}{\tfrac {\varphi }{2}}\sin \varphi }$$

the unit normal is

$${\displaystyle {\vec {n}}(\varphi )=(-\sin {\tfrac {3}{2}}\varphi ,\cos {\tfrac {3}{2}}\varphi )}$$

and the radius of curvature

$${\displaystyle \rho (\varphi )={\tfrac {8}{3}}a\sin {\tfrac {\varphi }{2}}\,.}$$

Hence the parametric equations of the evolute are

$${\displaystyle X(\varphi )=4a\sin ^{2}{\tfrac {\varphi }{2}}\cos \varphi -{\tfrac {8}{3}}a\sin {\tfrac {\varphi }{2}}\cdot \sin {\tfrac {3}{2}}\varphi =\cdots ={\tfrac {4}{3}}a\cos ^{2}{\tfrac {\varphi }{2}}\cos \varphi -{\tfrac {4}{3}}a\,,}$$

$${\displaystyle Y(\varphi )=4a\sin ^{2}{\tfrac {\varphi }{2}}\sin \varphi +{\tfrac {8}{3}}a\sin {\tfrac {\varphi }{2}}\cdot \cos {\tfrac {3}{2}}\varphi =\cdots ={\tfrac {4}{3}}a\cos ^{2}{\tfrac {\varphi }{2}}\sin \varphi \,.}$$

These equations describe a cardioid a third as large, rotated 180 degrees and shifted along the x-axis by ${\displaystyle -{\tfrac {4}{3}}a}$.

(Trigonometric formulae were used: ${\displaystyle \sin {\tfrac {3}{2}}\varphi =\sin {\tfrac {\varphi }{2}}\cos \varphi +\cos {\tfrac {\varphi }{2}}\sin \varphi \ ,\ \cos {\tfrac {3}{2}}\varphi =\cdots ,\ \sin \varphi =2\sin {\tfrac {\varphi }{2}}\cos {\tfrac {\varphi }{2}},\ \cos \varphi =\cdots \ .}$)

Proof

For a curve given in polar coordinates by a function ${\displaystyle r(\varphi )}$ the following connection to Cartesian coordinates hold:

$${\displaystyle {\begin{aligned}x(\varphi )&=r(\varphi )\cos \varphi \,,\\y(\varphi )&=r(\varphi )\sin \varphi \end{aligned}}}$$

and for the derivatives

$${\displaystyle {\begin{aligned}{\frac {dx}{d\varphi }}&=r'(\varphi )\cos \varphi -r(\varphi )\sin \varphi \,,\\{\frac {dy}{d\varphi }}&=r'(\varphi )\sin \varphi +r(\varphi )\cos \varphi \,.\end{aligned}}}$$

Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point ${\displaystyle (r(\varphi ),\varphi )}$:

$${\displaystyle {\frac {dy}{dx}}={\frac {r'(\varphi )\sin \varphi +r(\varphi )\cos \varphi }{r'(\varphi )\cos \varphi -r(\varphi )\sin \varphi }}.}$$

For the cardioids with the equations ${\displaystyle r=2a(1-\cos \varphi )\;}$ and ${\displaystyle r=2b(1+\cos \varphi )\ }$ respectively one gets:

$${\displaystyle {\frac {dy_{a}}{dx}}={\frac {\cos(\varphi )-\cos(2\varphi )}{\sin(2\varphi )-\sin(\varphi )}}}$$

and

$${\displaystyle {\frac {dy_{b}}{dx}}=-{\frac {\cos(\varphi )+\cos(2\varphi )}{\sin(2\varphi )+\sin(\varphi )}}\ .}$$

(The slope of any curve depends on ${\displaystyle \varphi }$ only, and not on the parameters ${\displaystyle a}$ or ${\displaystyle b}$!)

Hence

$${\displaystyle {\frac {dy_{a}}{dx}}\cdot {\frac {dy_{b}}{dx}}=\cdots =-{\frac {\cos ^{2}\varphi -\cos ^{2}(2\varphi )}{\sin ^{2}(2\varphi )-\sin ^{2}\varphi }}=-{\frac {-1+\cos ^{2}\varphi +1-\cos ^{2}2\varphi }{\sin ^{2}(2\varphi )-\sin ^{2}(\varphi )}}=-1\,.}$$

That means: Any curve of the first pencil intersects any curve of the second pencil orthogonally.

4 cardioids in polar representation and their position in the coordinate system