Polar Equation: Conversion Between Rectangular Form

Step 1: Substitute for x and y in x² + y² = 16 and solve for r.

$x=r\cos \theta$, $y=r\sin \theta$

x² + y² = 16

${\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2=16$ Sub

Solve for r

$r^2{\cos }^2\theta +r^2{\sin }^2\theta =16$ Square

$r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=16$ Factor r²

r²(1) = 16 Trig id.

r² = 16      Multiply

r = 4      Take root

Step 2: Substitute for x and y in x = 6 and solve for r.

$x=r\cos \theta$, $y=r\sin \theta$

x = 6

$r\cos \theta =6$ Sub

Solve for r

$r=\frac{6}{\cos \theta }$     $÷$ by $\cos \theta$

$r=6\sec \theta$ Simplify

Step 1: Square both sides of r = 5 and substitute for r².

$r^2=x^2+y^2$

r = 5

$r^2=5^2=25$    Square

$x^2+y^2=25$    Sub

Step 2: Determine the value of $tan \theta$ and equate this to $\frac{y}{x}$.

$\tan \theta =\frac{y}{x}$

$\theta =\frac{\pi }{6}$

$\tan \frac{\pi }{6}=\frac{\sqrt{3}}{3}$    Find tan θ

$\frac{\sqrt{3}}{3}=\frac{y}{x}$        Sub

$y=\frac{\sqrt{3}x}{3}$