Polar equation of a line

TEMPORARY PAGE - I'm working it out now so may as well use the steps

Line

• $y=mx+c$

Parametric

• $$r=\sqrt{t^2(m^2+1)+2mtc+c^2}$$
• $$\theta=\arctan\left(m+\frac{c}{t}\right)$$

Polar form

Method:

• Substitute:
  • $$t=\frac{c}{\tan(\theta)-m}$$ into $$r=\sqrt{t^2(m^2+1)+2mtc+c^2}$$

Working:

• $$r=\sqrt{\left(\frac{c}{\tan(\theta)-m}\right)^2(m^2+1)+2mc\left(\frac{c}{\tan(\theta)-m}\right)+c^2}$$
• $$r=|c|\sqrt{\frac{m^2+1}{(\tan(\theta)-m)^2}+\frac{2m}{\tan(\theta)-m}+1}$$
  • Solving the Quadratic equation (treating $\frac{1}{\tan(\theta)-m}$ as the variable we are quadratic in we see)
    • $a=m^2+1$
    • $b=2m$
    • $c=1$
  • To get roots $\frac{\pm j-m}{m^2+1}$ where $j=\sqrt{-1}$
  • noting that $m^2+1=(m-j)(m+j)$ we see that the roots are simply $\frac{-1}{m\pm j}$
  • this means $$(x-r_1)(x-r_2) = 0$$ whenever $x=r_1$ or $x=r_2$, so:
• $$r=|c|\sqrt{\left( \frac{1}{\tan(\theta)-m} + \frac{1}{m-j} \right)\left( \frac{1}{\tan(\theta)-m} + \frac{1}{m+j} \right)}$$

Questions

• Is this the best form?
• if $m=1$ and $c=1$ then
  • - this can't be the only special case!