Exercise 10.3.17

Proof.

(a)

As $\mathit{QR}=1$, where $Q\in C$ and $R$ is on the line $(\mathit{PQ})$, $R\in l$ is by definition on the limaçon determined by $P$ and $C$.

(b)

Let $l_1:y=\mathit{mx}$ be any nonvertical line passing through $P=(0,0)$, with $m=\tan <!--\; FUNCTION\; APPLICATION\; -->𝜃,-\pi ∕2<𝜃<\pi ∕2$, so the measure of the angle between the $x$-axis and $l_1$ is $𝜃$. Let $Q=(x_Q,y_Q),Q\ne P$, be the intersection point of $l_1$ with $C$. Then $x_Q\ne 0$.

$(x_Q,y_Q)$ is solution of the system

$$\begin{array}{llll}\hfill y& =\mathit{mx},& \hfill & \\ \hfill a^2& ={(x-a)}^2+y^2,& \hfill & \end{array}$$

which gives

$$y_Q=m{x}_Q,0=x_Q[(m^2+1)x_Q-2a],$$

with $x_Q\ne 0$, so

$$Q=\left(\frac{2a}{m^2+1},\frac{2\mathit{am}}{m^2+1}\right).$$

Since $m=\tan <!--\; FUNCTION\; APPLICATION\; -->𝜃$,

$$Q=(2a{\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃,2a\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃)=(a(\cos <!--\; FUNCTION\; APPLICATION\; -->(2𝜃)+1),a\sin <!--\; FUNCTION\; APPLICATION\; -->(2𝜃)).$$

Let $R=(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃,r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃)$ any point of $l_1$. Then $R$ is on the limaçon determined by $C$ and $P$ if and only if $\mathit{QR}=1$ (by continuity, the points at distance 1 on the vertical tangent to the circle at point $P$ are considered to be on the limaçon):

$$\begin{array}{llll}\hfill \mathit{QR}=1& ⟺1={(r\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃-2a{\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃)}^2+{(r\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃-2a\sin <!--\; FUNCTION\; APPLICATION\; -->𝜃\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃)}^2& \hfill & \\ \hfill & ⟺1=({\cos <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃+{\sin <!--\; FUNCTION\; APPLICATION\; -->}^2𝜃){(r-2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃)}^2& \hfill & \\ \hfill & ⟺1={(r-2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃)}^2& \hfill & \\ \hfill & ⟺r=1+2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃\mathrm{or}r=-1+2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃& \hfill & \end{array}$$

This can be interpreted geometrically. Let $P,A$ the intersection points of $C$ with the $x$-axis. As $\mathit{PA}$ is a diameter of $C$, the angle $\mathit{\angle PQA}$ is a right angle, so $\mathit{PQ}=2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃$. Therefore $\mathit{PR}=|\mathit{PQ}\pm 1|=|\pm 1+2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃|$.

The limaçon is the union of the curves $C_1,C_2$ with polar equations

As $\cos <!--\; FUNCTION\; APPLICATION\; -->(𝜃+\pi )=-\cos <!--\; FUNCTION\; APPLICATION\; -->(𝜃)$, if the point $R$ with polar coordinates $(r,𝜃)$ is on $C_1$, then $(-r,𝜃+\pi )$ is on $C_2$. But $(r,𝜃)$ and $(-r,𝜃+\pi )$ are polar coordinates of the same point! Therefore the two curves are identical if we let $𝜃$ vary in $ℝ$, and we obtain the complete curve if we let $𝜃$ vary in $]-\pi ,\pi ]$ in the equation of $C_1$.

The limaçon determined by $P$ and $C$ has polar equation

(c) Let $L$ the limaçon. By part (b), if $x\ne 0$

The two exceptional points $(0,\pm 1)$ satisfy this equation, so the Cartesian equation of the limaçon is

(We can also obtain this equation from the polar equation $r=1+2a\cos <!--\; FUNCTION\; APPLICATION\; -->𝜃$.) □