Why Convert a Cartesian Equation to Polar Form?
A Cartesian equation describes a curve using \(x\) and \(y\). A polar equation describes it using the distance \(r\) from the pole and the angle \(\theta\). Converting an equation between these forms is useful when a curve’s relationship to the pole or its direction from the pole is easier to express with polar variables.
The basic method is substitution. For a point with polar coordinates \((r,\theta)\), the Cartesian coordinates are \(x=r\cos\theta\) and \(y=r\sin\theta\). The squared distance from the origin is \(x^2+y^2=r^2\). These identities turn Cartesian equations into equations involving \(r\) and \(\theta\).
The condition \(x\ne 0\) matters: the tangent ratio is not defined at points where \(x=0\). For conversions that must include every point of a line, the substitution formulas \(x=r\cos\theta\) and \(y=r\sin\theta\) are safer than relying on a tangent equation alone.
Converting a Line
A line can be written in several Cartesian forms. For a line \(Ax+By=C\), where \(A\) and \(B\) are not both zero, substitute \(x=r\cos\theta\) and \(y=r\sin\theta\). The result is \(r(A\cos\theta+B\sin\theta)=C\). This form also works for vertical lines, which cannot be written as \(y=mx+b\).
When a line is written as \(y=mx+b\), the tangent identity can provide another route, as long as \(x\ne 0\). Substituting \(y/x=\tan\theta\) and \(x=r\cos\theta\) gives \(\tan\theta=m+\frac{b}{r\cos\theta}\). Multiplying by \(r\cos\theta\) leads to \(r\sin\theta=mr\cos\theta+b\). This is the same result as substituting directly into \(y=mx+b\).
The multiplied equation is often preferable to the tangent equation: it avoids leaving a quotient in the final expression and can include points with \(x=0\), provided the equation is checked in its original, undivided form.
Worked Example: Convert a Slope-Intercept Line
Convert \(y=2x+3\) to polar form. First use the tangent relation at points where \(x\ne 0\):
Now substitute \(x=r\cos\theta\):
Multiplying both sides by \(r\cos\theta\) gives
Thus one polar equation is \(r(\sin\theta-2\cos\theta)=3\). Check it using the Cartesian point \((0,3)\), which lies on the original line because \(3=2(0)+3\). Its polar values can be taken as \(r=3\), \(\theta=\frac{\pi}{2}\). Substitution gives \(3(1-2\cdot 0)=3\), so the point satisfies the polar equation as well.
Worked Example: Convert a Line in General Form
Convert \(3x-4y=12\) to polar form. Substitute the Cartesian expressions for \(x\) and \(y\):
To check, the Cartesian point \((4,0)\) lies on the line because \(3(4)-4(0)=12\). It has polar values \(r=4\), \(\theta=0\). In the converted equation, the left-hand side is \(4(3\cdot 1-4\cdot 0)=12\), matching the right-hand side.
Converting a Circle
The equation of a circle centered at \((a,b)\) with radius \(R\) is \((x-a)^2+(y-b)^2=R^2\), where \(R\ge 0\). Expanding the squares and using \(x^2+y^2=r^2\) gives a polar equation directly:
Equivalently, the equation can be kept in a form that shows the center and radius:
Expanding is often convenient for calculations. Keeping the squares can make it easier to see which Cartesian circle is being represented. A circle centered at the pole is a particularly direct case: if \(a=b=0\), then \(r^2=R^2\), and because \(r\ge 0\), this is \(r=R\).
Worked Example: Convert a Circle Centered at the Pole
Convert \(x^2+y^2=25\) to polar form. Apply the identity \(x^2+y^2=r^2\):
Since \(r\ge 0\), the equation is \(r=5\). Check with the Cartesian point \((3,4)\). It lies on the original circle because \(3^2+4^2=9+16=25\). Its distance from the pole is \(r=\sqrt{3^2+4^2}=5\), so it satisfies the polar equation.
Worked Example: Convert a Circle with an Off-Pole Center
Convert \((x-3)^2+(y+4)^2=16\) to polar form. First expand each square:
Now substitute \(x^2+y^2=r^2\), \(x=r\cos\theta\), and \(y=r\sin\theta\):
Check using \((3,0)\). Its distance from the center \((3,-4)\) is \(4\), so it lies on the original circle. Its polar values are \(r=3\) and \(\theta=0\). Substitution gives \(3^2-6(3)(1)+8(3)(0)+9=9-18+0+9=0\). The converted equation is satisfied.
Why Substitution Preserves the Curve
The conversion works because polar and Cartesian coordinates describe the same point in two forms. The algebra does not change the set of points when each Cartesian coordinate is replaced by its equal polar expression.
A point \((x,y)\) belongs to the curve precisely when its coordinates satisfy that equation.
Use \(x=r\cos\theta\) and \(y=r\sin\theta\), and use \(x^2+y^2=r^2\) wherever that combination occurs.
Because each replacement equals the original expression, the resulting equation holds exactly when the Cartesian equation holds for that point.
For example, the circle conversion follows by expanding its Cartesian equation and making these substitutions. Each line of algebra is an equality, so the final polar equation is satisfied by precisely the same points described in polar coordinates. The same reasoning applies to the line conversion.
There is one caution in this reasoning: an operation such as division can restrict which points are considered. Dividing by \(x\) requires \(x\ne 0\); dividing by \(r\cos\theta\) has the same restriction. Multiplying an equation by an expression is reversible only where that expression is nonzero. For that reason, use direct substitution for a general conversion, and treat equations obtained by division as applying only under their stated conditions.
A Common Mistake: Losing the Cosine Factor
When converting \(y=2x+3\), a tempting but incorrect step is to write \(\frac{y}{x}=2+\frac{3}{r}\). This incorrectly treats \(x\) as \(r\). In polar coordinates, \(x=r\cos\theta\), not \(r\).
The correction is to substitute the full expression for \(x\):
Alternatively, avoid the quotient and substitute directly into the original line:
This direct equation is easier to check and does not begin by excluding points with \(x=0\).
A Reliable Conversion Checklist
For a Cartesian line or circle, the following process keeps the algebra organized:
- Identify whether the equation is a line or circle and choose a useful Cartesian form.
- Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\).
- Replace \(x^2+y^2\) by \(r^2\) when it appears.
- Simplify without dividing by an expression that could be zero.
- Check the result with a point known to satisfy the original equation.
The most important distinction is between substitution and division. Substitution replaces each coordinate by an equal expression. Division can discard points, so any equation derived by dividing must retain the condition that the divisor is nonzero.
Check Your Understanding
Work through these questions by making the substitutions before simplifying.
- Convert \(y=-x+2\) into a polar equation using direct substitution.
- Convert \(5x+2y=7\) into a polar equation.
- Convert \(x^2+y^2=49\) into polar form and state why \(r\) is nonnegative.
- Convert \((x-1)^2+(y-2)^2=9\) into a polar equation by expanding first.
- Why must the tangent relation be used with the condition \(x\ne 0\)?
