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6.6: Rectangular to Polar Form for Equations

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
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Practice Rectangular to Polar Form for Equations
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Practice Now

You are working diligently in your math class when your teacher gives you an equation to graph:

\begin{align*}(x + 1)^2 - (y + 2)^2 = 7\end{align*}(x+1)2(y+2)2=7

As you start to consider how to rearrange this equation, you are told that the goal of the class is to convert the equation to polar form instead of rectangular form.

Can you find a way to do this?

By the end of this Concept, you'll be able to convert this equation to polar form.

Watch This

James Sousa Example: Find the Polar Equation for a Line

Guidance

Interestingly, a rectangular coordinate system isn't the only way to plot values. A polar system can be useful. However, it will often be the case that there are one or more equations that need to be converted from rectangular to polar form. To write a rectangular equation in polar form, the conversion equations of \begin{align*}x = r \cos \theta\end{align*}x=rcosθ and \begin{align*}y = r \sin \theta\end{align*}y=rsinθ are used.

If the graph of the polar equation is the same as the graph of the rectangular equation, then the conversion has been determined correctly.

\begin{align*}(x-2)^2+y^2=4\end{align*}(x2)2+y2=4

The rectangular equation \begin{align*}(x - 2)^2 + y^2 = 4\end{align*}(x2)2+y2=4 represents a circle with center (2, 0) and a radius of 2 units. The polar equation \begin{align*}r = 4 \cos \theta\end{align*}r=4cosθ is a circle with center (2, 0) and a radius of 2 units.

Example A

Write the rectangular equation \begin{align*}x^2 + y^2 = 2x\end{align*}x2+y2=2x in polar form.

Solution: Remember \begin{align*}r = \sqrt{x^2 + y^2}, r^2 = x^2 + y^2\end{align*}r=x2+y2,r2=x2+y2 and \begin{align*}x = r \cos \theta\end{align*}x=rcosθ.

\begin{align*}x^2 + y^2 &= 2x\\ r^2 &= 2(r \cos \theta) && Pythagorean \ Theorem \ and \ x = r \cos \theta\\ r^2 &= 2r \cos \theta\\ r &= 2 \cos \theta && Divide \ each \ side \ by \ r\end{align*}x2+y2r2r2r=2x=2(rcosθ)=2rcosθ=2cosθPythagorean Theorem and x=rcosθDivide each side by r

Example B

Write \begin{align*}(x - 2)^2 + y^2 = 4\end{align*}(x2)2+y2=4 in polar form.

Remember \begin{align*}x = r \cos \theta\end{align*}x=rcosθ and \begin{align*}y = r \sin \theta\end{align*}y=rsinθ.

\begin{align*}&(x - 2)^2 + y^2 = 4\\ &(r \cos \theta - 2)^2 + (r \sin \theta)^2 = 4 && x = r \cos \theta \ and \ y = r \sin \theta\\ &r^2 \cos^2 \theta - 4r \cos \theta + 4 + r^2 \sin^2 \theta = 4 && expand \ the \ terms\\ &r^2 \cos^2 \theta - 4r \cos \theta + r^2 \sin^2 \theta = 0 && subtract \ 4 \ from \ each \ side\\ &r^2 \cos^2 \theta + r^2 \sin^2 \theta = 4r \cos \theta && isolate \ the \ squared \ terms\\ &r^2 (\cos^2 \theta + \sin^2 \theta) = 4r \cos \theta && factor \ r^2 - a \ common \ factor\\ &r^2 = 4r \cos \theta && Pythagorean \ Identity\\ &r = 4 \cos \theta && Divide \ each \ side \ by \ r\end{align*}(x2)2+y2=4(rcosθ2)2+(rsinθ)2=4r2cos2θ4rcosθ+4+r2sin2θ=4r2cos2θ4rcosθ+r2sin2θ=0r2cos2θ+r2sin2θ=4rcosθr2(cos2θ+sin2θ)=4rcosθr2=4rcosθr=4cosθx=rcosθ and y=rsinθexpand the termssubtract 4 from each sideisolate the squared termsfactor r2a common factorPythagorean IdentityDivide each side by r

Example C

Write the rectangular equation \begin{align*}(x+4)^2 + (y-1)^2 = 17\end{align*}(x+4)2+(y1)2=17 in polar form.

\begin{align*}&(x+4)^2 + (y-1)^2 = 17\\ &(r \cos \theta + 4)^2 + (r \sin \theta - 1)^2 = 17 && x = r \cos \theta \ and \ y = r \sin \theta\\ &r^2 \cos^2 \theta + 8r \cos \theta + 16 + r^2 \sin^2 \theta - 2r \sin \theta + 1 = 17 && expand \ the \ terms\\ &r^2 \cos^2 \theta + 8r \cos \theta - 2r \sin \theta + r^2 \sin^2 \theta = 0 && subtract \ 17 \ from \ each \ side\\ &r^2 \cos^2 \theta + r^2 \sin^2 \theta = -8r \cos \theta + 2r \sin \theta && isolate \ the \ squared \ terms\\ &r^2 (\cos^2 \theta + \sin^2 \theta) = -2r (4\cos \theta - \sin \theta) && factor \ r^2 - a \ common \ factor\\ &r^2 = -2r (4\cos \theta - \sin \theta) && Pythagorean \ Identity\\ &r = -2(4\cos \theta - \sin \theta) && Divide \ each \ side \ by \ r\end{align*}(x+4)2+(y1)2=17(rcosθ+4)2+(rsinθ1)2=17r2cos2θ+8rcosθ+16+r2sin2θ2rsinθ+1=17r2cos2θ+8rcosθ2rsinθ+r2sin2θ=0r2cos2θ+r2sin2θ=8rcosθ+2rsinθr2(cos2θ+sin2θ)=2r(4cosθsinθ)r2=2r(4cosθsinθ)r=2(4cosθsinθ)x=rcosθ and y=rsinθexpand the termssubtract 17 from each sideisolate the squared termsfactor r2a common factorPythagorean IdentityDivide each side by r

Vocabulary

Polar Coordinates: A set of polar coordinates are a set of coordinates plotted on a system that uses the distance from the origin and angle from an axis to describe location.

Rectangular Coordinates: A set of rectangular coordinates are a set of coordinates plotted on a system using basis axes at right angles to each other.

Guided Practice

1. Write the rectangular equation \begin{align*}(x - 4)^2 + (y - 3)^2 = 25\end{align*}(x4)2+(y3)2=25 in polar form.

2. Write the rectangular equation \begin{align*}3x - 2y = 1\end{align*}3x2y=1 in polar form.

3. Write the rectangular equation \begin{align*}x^2 + y^2 - 4x + 2y = 0\end{align*}x2+y24x+2y=0 in polar form.

Solutions:

1.

\begin{align*}(x - 4)^2 + (y - 3)^2 & = 25 \\ x^2 - 8x + 16 + y^2 - 6y + 9 & = 25 \\ x^2 - 8x + y^2 - 6y + 25 & = 25 \\ x^2 - 8x + y^2 - 6y & = 0 \\ x^2 + y^2 - 8x - 6y & = 0 \\ r^2 - 8(r \cos \theta) - 6(r \sin \theta) & = 0 \\ r^2 - 8r \cos \theta - 6r \sin \theta & = 0 \\ r(r - 8 \cos \theta - 6 \sin \theta) & = 0 \\ r = 0\ \text{or}\ r - 8 \cos \theta - 6 \sin \theta & = 0 \\ r = 0\ \text{or}\ r & = 8 \cos \theta + 6 \sin \theta\end{align*}(x4)2+(y3)2x28x+16+y26y+9x28x+y26y+25x28x+y26yx2+y28x6yr28(rcosθ)6(rsinθ)r28rcosθ6rsinθr(r8cosθ6sinθ)r=0 or r8cosθ6sinθr=0 or r=25=25=25=0=0=0=0=0=0=8cosθ+6sinθ

From graphing \begin{align*}r-8\cos \theta -6\sin \theta =0\end{align*}r8cosθ6sinθ=0, we see that the additional solutions are 0 and 8.

2.

\begin{align*}3x - 2y & = 1 \\ 3r \cos \theta - 2r \sin \theta & = 1 \\ r (3 \cos \theta - 2 \sin \theta) & = 1 \\ r & = \frac{1}{3 \cos \theta - 2 \sin \theta}\end{align*}3x2y3rcosθ2rsinθr(3cosθ2sinθ)r=1=1=1=13cosθ2sinθ

3.

\begin{align*}x^2 + y^2 - 4x + 2y & = 0 \\ r^2 \cos^2 \theta + r^2 \sin^2 \theta - 4 r \cos \theta + 2 r \sin \theta & = 0 \\ r^2 (\sin^2 \theta + \cos^2 \theta) - 4 r \cos \theta + 2 r \sin \theta & = 0 \\ r (r - 4 \cos \theta + 2 \sin \theta) & = 0 \\ r = 0\ \text{or}\ r - 4 \cos \theta + 2 \sin \theta & = 0 \\ r = 0\ \text{or}\ r & = 4 \cos \theta - 2 \sin \theta\end{align*}x2+y24x+2yr2cos2θ+r2sin2θ4rcosθ+2rsinθr2(sin2θ+cos2θ)4rcosθ+2rsinθr(r4cosθ+2sinθ)r=0 or r4cosθ+2sinθr=0 or r=0=0=0=0=0=4cosθ2sinθ

Concept Problem Solution

The original equation to convert is:

\begin{align*}(x + 1)^2 - (y + 2)^2 = 7\end{align*}(x+1)2(y+2)2=7

You can substitute \begin{align*}x = r\cos \theta\end{align*}x=rcosθ and \begin{align*}y = r\sin \theta\end{align*}y=rsinθ into the equation, and then simplify:

\begin{align*} (r\cos \theta +1)^2 - (r\sin \theta + 2)^2 = 7\\ (r^2\cos^2 \theta + 2r\cos \theta +1) - (r^2\sin^2 \theta + 2r\cos \theta + 4) = 7\\ r^2(\cos^2 \theta + \sin^2 \theta) + 4r\cos \theta + 5 = 7\\ r^2 + 4r\cos \theta = 2 \end{align*}(rcosθ+1)2(rsinθ+2)2=7(r2cos2θ+2rcosθ+1)(r2sin2θ+2rcosθ+4)=7r2(cos2θ+sin2θ)+4rcosθ+5=7r2+4rcosθ=2

Practice

Write each rectangular equation in polar form.

  1. \begin{align*}x=3\end{align*}x=3
  2. \begin{align*}y=4\end{align*}y=4
  3. \begin{align*}x^2+y^2=4\end{align*}x2+y2=4
  4. \begin{align*}x^2+y^2=9\end{align*}x2+y2=9
  5. \begin{align*}(x-1)^2+y^2=1\end{align*}(x1)2+y2=1
  6. \begin{align*}(x-2)^2+(y-3)^2=13\end{align*}(x2)2+(y3)2=13
  7. \begin{align*}(x-1)^2+(y-3)^2=10\end{align*}(x1)2+(y3)2=10
  8. \begin{align*}(x+2)^2+(y+2)^2=8\end{align*}(x+2)2+(y+2)2=8
  9. \begin{align*}(x+5)^2+(y-1)^2=26\end{align*}
  10. \begin{align*}x^2+(y-6)^2=36\end{align*}
  11. \begin{align*}x^2+(y+2)^2=4\end{align*}
  12. \begin{align*}2x+5y=11\end{align*}
  13. \begin{align*}4x-7y=10\end{align*}
  14. \begin{align*}x+5y=8\end{align*}
  15. \begin{align*}3x-4y=15\end{align*}

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Date Created:
Sep 26, 2012
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Mar 23, 2016
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