Why Polar Intersections Need Care
In Cartesian coordinates, a point has one pair of coordinates. In polar coordinates, a point can have many representations. In particular, a negative radius points in the direction opposite to its angle. As a result, two polar curves can meet even when their radius functions are not equal at the same angle.
To find all intersections, we must compare both curves’ representations of a point. For nonzero points, this means checking equal angles and angles that differ by \(\pi\). The pole requires a separate check, because it is represented by radius zero at every angle.
How Polar Representations Describe the Same Point
A polar pair \((r,\theta)\) represents the Cartesian point \((r\cos\theta,r\sin\theta)\). For a nonzero point, increasing the angle by \(2\pi\) leaves the point unchanged. Reversing the sign of the radius and adding \(\pi\) to the angle also leaves it unchanged:
For example, \((1,0)\) and \((-1,\pi)\) both represent the point \((1,0)\). This is why comparing only the radii at the same angle can overlook a shared point.
Suppose two polar curves are given by \(r=f(\theta)\) and \(r=g(\theta)\), each on an angular interval that traces the curve being considered. If the functions are periodic and a complete trace is taken over a full revolution, they can be treated as defined for all real angles by periodic extension.
The Two Equations for Nonzero Intersections
First compare the curves at the same angle. A solution of \(f(\theta)=g(\theta)\) gives the same signed radius in the same direction, so it gives a common point.
Next compare the first curve at angle \(\theta\) with the second curve at angle \(\theta+\pi\). The latter point has Cartesian coordinates
It therefore agrees with the first curve’s point at angle \(\theta\) exactly when \(f(\theta)=-g(\theta+\pi)\). For complete periodic traces, these are the two equations to solve for nonzero intersections:
Any solutions must be checked against the curves’ parameter ranges. When working over a full revolution, angles are considered modulo \(2\pi\), and solutions that represent the same point are recorded only once.
Why These Checks Are Complete
The two equations are not merely useful tests: together they cover every possible nonzero intersection. The reason is that any nonzero point has a definite direction from the pole once its radius is made positive. The same point on two curves must therefore have either the same direction or directions differing by \(\pi\), depending on the signs of the signed radii.
At parameter values \(\alpha\) and \(\beta\), the curves give \(f(\alpha)(\cos\alpha,\sin\alpha)\) and \(g(\beta)(\cos\beta,\sin\beta)\).
For a nonzero common point, after accounting for signed radii, the angles must agree modulo \(2\pi\) or differ by \(\pi\) modulo \(2\pi\).
Equal angles require equal signed radii, giving \(f(\theta)=g(\theta)\). Opposite angles require opposite signed radii, giving \(f(\theta)=-g(\theta+\pi)\).
If either radius is zero, its angle does not determine a direction. The origin is common precisely when each curve reaches radius zero somewhere in its allowed parameter range.
This argument assumes the parameter values used are allowed on both curves, with periodic extension only when the curves are being treated as complete periodic traces. It also explains why a pole intersection cannot always be found by comparing radii at a single shared angle.
Worked Examples
Worked Example: A Cardioid and a Circle
Find all intersections of \(r=2+2\cos\theta\) and \(r=2\), using a full revolution.
Equal-angle check. Set the radii equal:
Over \(0\leq\theta<2\pi\), the solutions are \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\). Both radii are \(2\). Their Cartesian points are
Each point satisfies both curve equations: at the listed angle, the cardioid radius is \(2+2(0)=2\), equal to the circle radius.
Opposite-angle check. The equation is
This would require \(\cos\theta=-2\), which is impossible because \(-1\leq\cos\theta\leq1\). There are no additional intersections from opposite-angle representations.
Pole check. The cardioid reaches radius zero when \(2+2\cos\theta=0\), or \(\cos\theta=-1\). The circle \(r=2\) never has radius zero, so the pole is not a common point. Thus the complete intersection set is \((0,2)\) and \((0,-2)\).
Worked Example: An Intersection Missed by Equal Angles
Find all intersections of \(r=1+2\cos\theta\) and \(r=1\), again using a full revolution.
Equal-angle check. Solving \(1+2\cos\theta=1\) gives
The solutions are \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\). The shared radius at each angle is \(1\), giving the points \((0,1)\) and \((0,-1)\). Substitution into the first curve gives \(1+2(0)=1\), and into the second gives radius \(1\), as required.
Opposite-angle check. Since the second curve has \(g(\theta+\pi)=1\), solve
The solution is \(\theta=\pi\). The first curve has radius \(1+2\cos\pi=-1\) there. Its Cartesian point is
The second curve represents the same point at angle \(0\), since its radius is \(1\) and \((1\cos0,1\sin0)=(1,0)\). This verifies the additional intersection.
Pole check. The first curve reaches radius zero when \(\cos\theta=-\frac12\), but the second curve never has radius zero. The pole is not common. The full intersection set is \((0,1)\), \((0,-1)\), and \((1,0)\).
Worked Example: Two Curves with a Pole Intersection
Find the intersections of \(r=1+\cos\theta\) and \(r=1+\sin\theta\) over a full revolution.
Equal-angle check. Equating the radii gives
In \(0\leq\theta<2\pi\), the solutions are \(\frac{\pi}{4}\) and \(\frac{5\pi}{4}\). At \(\theta=\frac{\pi}{4}\), the common radius is \(1+\frac{\sqrt2}{2}\), so the point is
Both radii equal \(1+\frac{\sqrt2}{2}\) at \(\frac{\pi}{4}\), verifying that both curves represent this point.
At \(\theta=\frac{5\pi}{4}\), the common radius is \(1-\frac{\sqrt2}{2}\). Since \(\cos\frac{5\pi}{4}=\sin\frac{5\pi}{4}=-\frac{\sqrt2}{2}\), the Cartesian point is
Substituting the angle into either radius function gives \(1-\frac{\sqrt2}{2}\), so both curves represent this point.
Opposite-angle check. Here \(g(\theta+\pi)=1+\sin(\theta+\pi)=1-\sin\theta\). Thus
But \(\cos\theta-\sin\theta=\sqrt2\cos(\theta+\frac{\pi}{4})\), whose values lie between \(-\sqrt2\) and \(\sqrt2\). Since \(-2<-\sqrt2\), this equation has no solutions.
Pole check. The first curve has radius zero at \(\theta=\pi\), and the second has radius zero at \(\theta=\frac{3\pi}{2}\). Both therefore pass through \((0,0)\), even though the zero radii occur at different angles. The complete intersection set consists of the two points found by the equal-angle equation and the pole.
A Common Mistake: Solving Only at the Same Angle
A tempting shortcut is to solve \(f(\theta)=g(\theta)\) and assume every intersection has been found. The second worked example shows why this fails: equal-angle solutions produce \((0,1)\) and \((0,-1)\), but the curves also meet at \((1,0)\), represented by radius \(-1\) at angle \(\pi\) on the first curve and radius \(1\) at angle \(0\) on the second.
Finding the intersection points is a necessary step in many polar-curve problems. The list of intersections identifies angles or points where the relationship between curves may change. Determining which curve is farther from the pole on each interval is a separate task; intersections alone do not establish that ordering.
Check Your Understanding
Use the two intersection equations and remember to check the pole separately.
- Why can two polar representations with different angles describe the same nonzero point?
- For curves \(r=f(\theta)\) and \(r=g(\theta)\), what equation checks opposite-angle representations?
- What are the equal-angle intersections of \(r=2\) and \(r=2+2\sin\theta\) over a full revolution?
- Why is the pole not fully checked by solving \(f(\theta)=g(\theta)\)?
- For \(r=1+2\cos\theta\) and \(r=1\), which intersection is missed if only equal angles are checked?