Circles Offset from the Pole

From Polar Equations to Offset Circles

Some polar equations that appear to describe changing radius with angle actually trace familiar Cartesian curves. In particular, \(r=2a\cos\theta\) and \(r=2a\sin\theta\) each describe a circle, usually shifted away from the pole. Converting these equations makes the geometry visible: the Cartesian equation reveals the center, the radius, and the direction of the shift.

We will use the standard polar-to-Cartesian relations \(x=r\cos\theta\), \(y=r\sin\theta\), and \(r^2=x^2+y^2\). The parameter \(a\) is usually taken to be positive. We will also note what changes if \(a\) is zero or negative.

The Cosine Equation: A Circle on the Horizontal Axis

Start with \(r=2a\cos\theta\). Multiplying both sides by \(r\) gives \(r^2=2ar\cos\theta\). Substitute \(r^2=x^2+y^2\) and \(r\cos\theta=x\):

To identify the circle, complete the square in \(x\). Add \(a^2\) to both sides:

The standard equation of a circle with center \((h,k)\) and radius \(R\) is \((x-h)^2+(y-k)^2=R^2\). Therefore, when \(a>0\), this circle has center \((a,0)\) and radius \(a\). Its center lies on the positive horizontal axis, a distance \(a\) to the right of the pole.

The circle passes through the pole: substituting \((x,y)=(0,0)\) into \((x-a)^2+y^2=a^2\) gives \(a^2=a^2\). It is tangent to the vertical axis at the pole, since the center-to-pole radius is horizontal. Its horizontal extent is from \(x=0\) to \(x=2a\), and its vertical extent is from \(y=-a\) to \(y=a\).

The Sine Equation: A Circle on the Vertical Axis

Now start with \(r=2a\sin\theta\). Multiplying by \(r\) gives \(r^2=2ar\sin\theta\). Since \(r\sin\theta=y\), conversion gives:

Complete the square in \(y\):

For \(a>0\), this is a circle with center \((0,a)\) and radius \(a\). Its center lies on the positive vertical axis, a distance \(a\) above the pole. Like the cosine circle, it passes through the pole. The circle is tangent to the horizontal axis there, and its vertical extent runs from \(y=0\) to \(y=2a\).

Why the Conversion Gives the Same Curve

The algebra above shows how every point represented by either polar equation satisfies its Cartesian circle equation. The reverse implication matters too: it confirms that the Cartesian equation does not describe a different or larger curve.

For the cosine equation, suppose \((x,y)\) satisfies \((x-a)^2+y^2=a^2\). Expanding and cancelling \(a^2\) gives \(x^2+y^2=2ax\). For a polar representation of this point, \(x=r\cos\theta\) and \(x^2+y^2=r^2\), so \(r^2=2ar\cos\theta\). If \(r\ne 0\), divide by \(r\) to obtain \(r=2a\cos\theta\). If \(r=0\), the point is the pole, which the polar equation also represents at angles with \(\cos\theta=0\). Thus the polar and Cartesian equations describe the same locus. The sine case follows in the same way using \(y=r\sin\theta\).

Parameter Sign and Circle Orientation

The completed-square equations also clarify the role of the sign of \(a\). For any real \(a\), the cosine equation becomes \((x-a)^2+y^2=a^2\); the sine equation becomes \(x^2+(y-a)^2=a^2\). If \(a<0\), the radius is \(|a|\), not \(a\), because a radius is nonnegative. The cosine circle’s center is then \((a,0)\), on the negative horizontal axis; the sine circle’s center is \((0,a)\), on the negative vertical axis. If \(a=0\), either equation reduces to \(r=0\), representing only the pole, rather than a circle with positive radius.

For example, \(r=-4\sin\theta\) has \(2a=-4\), so \(a=-2\). Its Cartesian equation is \(x^2+y^2=-4y\), or \(x^2+(y+2)^2=4\). The center is \((0,-2)\) and the radius is \(2\): the circle is shifted downward. This sign interpretation concerns the Cartesian locus, not just whether a particular polar representation uses positive or negative radius.

A Common Conversion Mistake

A frequent error is to replace \(r^2\) by \(x^2+y^2\) correctly, but then replace \(r\cos\theta\) by \(x^2\), or to treat \(r\) itself as \(x\). The identities needed here are \(r\cos\theta=x\) and \(r\sin\theta=y\), not \(r\cos\theta=x^2\) or \(r=x\).

For instance, an incorrect conversion of \(r=2a\cos\theta\) might claim \(x^2+y^2=2ax^2\). That expression has not followed from the polar equation: its right-hand side should be \(2a(r\cos\theta)=2ax\). The corrected calculation is:

A reliable check is to identify each product before substituting: \(r^2=x^2+y^2\), \(r\cos\theta=x\), and \(r\sin\theta=y\). Then complete the square in the coordinate that appears linearly.

Comparing the Two Circles

The two equations differ only in which coordinate appears linearly: \(x\) for the cosine equation and \(y\) for the sine equation. That determines whether the center is shifted horizontally or vertically. Both circles have center-to-pole distance equal to their radius, which is why both pass through the pole and are tangent to an axis there.