Why Convert a Polar Equation?
A polar equation describes points using a distance \(r\) from the pole and an angle \(\theta\). A Cartesian equation describes the same points using coordinates \(x\) and \(y\). Converting between the two forms can make a curve easier to recognize algebraically, but each conversion step must preserve the points in the original equation.
The main tools are the coordinate identities. They let us replace trigonometric expressions involving \(r\) and \(\theta\) with expressions involving \(x\) and \(y\). The main caution is that operations such as squaring, dividing, or clearing a denominator can change which points satisfy an equation.
These identities do not depend on choosing a particular angle representative. In examples where \(r\) is used as a distance, take \(r\ge 0\). If a problem uses signed radii, preserve the sign information instead of replacing \(r\) by \(\sqrt{x^2+y^2}\) without checking.
A Reliable Conversion Method
First identify which parts of the polar equation can be replaced directly. For example, \(r\cos\theta\) becomes \(x\), \(r\sin\theta\) becomes \(y\), and \(r^2\) becomes \(x^2+y^2\). If the equation contains \(\cos\theta\) alone, then \(\cos\theta=x/r\) when \(r\ne 0\); that substitution introduces a denominator and requires attention to the case \(r=0\).
Sometimes multiplying through by \(r\) or by another expression removes a denominator. This can be useful, but multiplication can make an equation true where the factor being multiplied by is zero. Squaring can also discard the sign of an expression. Whenever either operation occurs, check the resulting equation against the original relation and any restrictions that came with it.
Look for \(r\cos\theta\), \(r\sin\theta\), \(r^2\), or trigonometric functions that can be written as \(x/r\) or \(y/r\).
Use the coordinate identities to form an equation involving \(x\) and \(y\).
Record any denominator restrictions and any sign conditions created by squaring.
Substitute a point into the original equation, or verify that the reverse substitutions recover it.
When the Substitution Is Direct
Equations with products such as \(r\cos\theta\) or \(r\sin\theta\) often convert cleanly. Equations involving \(r^2\) are also direct. A common pattern is to multiply an equation by \(r\), so that a standalone \(\cos\theta\) or \(\sin\theta\) appears as part of \(r\cos\theta\) or \(r\sin\theta\). The worked examples show how the resulting Cartesian equations can be checked.
Worked Example: Convert \(r=6\cos\theta\)
Multiply both sides by \(r\). This gives \(r^2=6r\cos\theta\). Now substitute \(r^2=x^2+y^2\) and \(r\cos\theta=x\):
Complete the square in \(x\):
This is a circle with center \((3,0)\) and radius \(3\). To check a point, take \((6,0)\). It has \(r=6\) and \(\theta=0\), so the original equation gives \(6=6\cos 0=6\). The Cartesian equation also gives \((6-3)^2+0^2=9\).
The origin is also consistent: in the original equation, \(r=0\) and \(\theta=\pi/2\) give \(0=6\cos(\pi/2)=0\); in the Cartesian equation, \((0-3)^2+0^2=9\). Thus multiplying by \(r\) has not added the origin as a new point in this example.
Worked Example: Convert \(r=4\sin\theta\)
Multiply by \(r\) and use \(r^2=x^2+y^2\) and \(r\sin\theta=y\):
Complete the square in \(y\):
The result is a circle with center \((0,2)\) and radius \(2\). Check the point \((0,4)\): its polar coordinates include \(r=4,\theta=\pi/2\), and the original equation gives \(4=4\sin(\pi/2)=4\). In Cartesian form, \(0^2+(4-2)^2=4\), as required.
For the reverse check, any point on this circle satisfies \(x^2+y^2=4y\). With \(r\ge0\), this says \(r^2=4r\sin\theta\). If \(r>0\), divide by \(r\) to recover \(r=4\sin\theta\). If \(r=0\), the point is the origin; the original equation holds for an angle with \(\sin\theta=0\). This checks both the nonzero-radius case and the pole.
Equations That Need an Extra Check
If a polar equation contains \(\cos\theta\) by itself, replacing it with \(x/r\) is valid only when \(r\ne0\). If it contains a denominator, the original denominator must be nonzero. A later algebraic equation may look simpler but still needs to retain the restrictions needed to reverse the steps.
Squaring is a particularly important case. If an equation has the form \(\sqrt{A}=B\), squaring gives \(A=B^2\), but that equation alone does not ensure \(B\ge0\). The original equation requires both \(A=B^2\) and \(B\ge0\). Without that sign condition, the squared equation can include points that fail the original one.
Worked Example: Convert a Polar Equation to a Parabola
Consider \(r=\dfrac{2}{1+\cos\theta}\), with the restriction \(1+\cos\theta\ne0\). Multiply by \(1+\cos\theta\), then use \(r\cos\theta=x\):
Using the nonnegative-distance convention, \(r=\sqrt{x^2+y^2}\). Therefore \(\sqrt{x^2+y^2}=2-x\). Squaring gives
The squaring step requires \(2-x\ge0\), or \(x\le2\). In this case, \(y^2=4-4x\) itself implies \(x\le1\), so the sign condition is automatically satisfied. The Cartesian equation is the parabola \(y^2=4-4x\).
Check \((0,2)\). It satisfies the Cartesian equation because \(2^2=4-4(0)=4\). Its polar coordinates include \(r=2,\theta=\pi/2\), and the original equation gives \(2=2/(1+0)=2\). For the reverse direction, a point on the parabola has \(r^2=x^2+y^2=(2-x)^2\); since \(x\le1\), \(2-x>0\), so \(r=2-x\). Then \(1+\cos\theta=1+x/r=(r+x)/r=2/r\), which is nonzero, and \(2/(1+\cos\theta)=r\). Thus the original equation is recovered.
Why Squaring Can Add Points
The need to keep a sign condition is not merely a formality. For example, with \(r\ge0\), consider the polar equation \(r=2r\cos\theta-1\). Since \(r\cos\theta=x\), this becomes \(\sqrt{x^2+y^2}=2x-1\). Its right-hand side must be nonnegative, so \(x\ge\frac12\). Squaring gives
The squared equation alone includes \((0,1)\), because \(1=3(0)^2-4(0)+1\). But the unsquared equation at that point reads \(1=-1\), which is false. The point is extraneous; it appeared because squaring erased the sign condition. The correct Cartesian description includes the restriction \(x\ge\frac12\). As a valid check, \((1,0)\) satisfies the unsquared equation: \(\sqrt{1}=2(1)-1=1\).
Why the Coordinate Identities Work
The substitutions are justified directly by the definitions of polar coordinates. A point at distance \(r\) and angle \(\theta\) has horizontal displacement \(r\cos\theta\) and vertical displacement \(r\sin\theta\), so \(x=r\cos\theta\) and \(y=r\sin\theta\). Squaring and adding yields
because \(\cos^2\theta+\sin^2\theta=1\). This also explains why \(r^2\) can always be replaced by \(x^2+y^2\), while replacing \(r\) itself by a square root requires a choice of sign convention.
Check Your Understanding
For each question, write the Cartesian equation and note any restriction needed to preserve the original equation.
- Convert \(r=8\cos\theta\) to Cartesian form and identify its center and radius.
- Convert \(r=6\sin\theta\) to Cartesian form and identify its center and radius.
- When converting an equation using \(\cos\theta=x/r\), what case must be checked separately?
- Why can squaring \(\sqrt{x^2+y^2}=2x-1\) introduce points that do not satisfy the original equation?
- For \(r=\dfrac{3}{1-\cos\theta}\), what denominator restriction must be considered before simplifying?
