Converting Off-Center Circles into Polar Bounds

Why an Off-Center Circle Needs Careful Polar Bounds

A circle centered at the origin has the same radial extent in every direction. Moving its center changes that: a ray from the origin may miss the disk, touch it once, or pass through it over a nonzero radial interval. To use polar coordinates accurately, we must determine both which angles meet the circle and which radii lie within it.

The key is to substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into the circle inequality, then interpret the resulting quadratic inequality in \(r\). The location of the origin relative to the circle determines the form of the bounds.

Write \(d=\sqrt{a^2+b^2}\) for the distance from the origin to the center. For an off-center circle, \(d>0\). Choose an angle \(\phi\) such that \(a=d\cos\phi\) and \(b=d\sin\phi\). Angles may be understood modulo \(2\pi\). Polar coordinates use \(r\ge 0\), so only nonnegative roots of the radial equation can describe points on a ray from the origin.

Substitute and Solve Along a Ray

Using \(x=r\cos\theta\), \(y=r\sin\theta\), the closed-disk inequality becomes

Expanding and collecting powers of \(r\) gives

For a fixed angle, this is a quadratic inequality in \(r\). It is useful to set \(q(\theta)=a\cos\theta+b\sin\theta=d\cos(\theta-\phi)\). The quadratic then has the form \(r^2-2q(\theta)r+d^2-R^2\le 0\), and its discriminant, after dividing the usual discriminant by four, is

This quantity tells us whether the line in direction \(\theta\) meets the circle. But the discriminant alone is not enough: the roots must also be interpreted with \(r\ge 0\). The following cases make that interpretation precise.

Three Positions of the Origin

Origin outside the disk: \(d>R\). A ray meets the closed disk only when it points sufficiently toward the center. The discriminant condition is \(|\sin(\theta-\phi)|\le R/d\). This condition alone also permits directions pointing away from the center, so we additionally need \(q(\theta)>0\) for a nonzero interval of nonnegative radii. Define \(\alpha=\arcsin(R/d)\), with \(0<\alpha<\pi/2\). The allowed angles for a nonzero intersection are \(\phi-\alpha\le\theta\le\phi+\alpha\), modulo \(2\pi\). At such an angle the radial roots are \(q(\theta)\pm\sqrt{R^2-d^2\sin^2(\theta-\phi)}\). The closed disk lies between these roots. At the two endpoint angles the roots coincide, giving a tangent point; the disk’s interior uses strict angular bounds and strict radial bounds.

Origin inside the disk: \(d<R\). The constant term \(d^2-R^2\) is negative, so the quadratic has one negative root and one positive root for every angle. The ray begins inside the disk and exits at the positive root. Every angle occurs, and the outer radial limit is \(q(\theta)+\sqrt{q(\theta)^2+R^2-d^2}\). The closed disk on that ray has \(0\le r\le r_+(\theta)\). Its interior includes the origin; for \(r>0\), points in the interior satisfy \(r<r_+(\theta)\).

Origin on the circle: \(d=R\). The radial inequality factors as \(r(r-2q(\theta))\le 0\). If \(q(\theta)>0\), the closed disk runs along the ray from \(r=0\) to \(r=2q(\theta)\). If \(q(\theta)=0\), the ray meets the disk only at the origin; if \(q(\theta)<0\), there is no positive-radius intersection. Since \(q(\theta)=d\cos(\theta-\phi)\), the nonzero radial intervals occur for \(\phi-\pi/2<\theta<\phi+\pi/2\), modulo \(2\pi\). The interior excludes both the origin and the outer circle, so on those rays it has \(0<r<2q(\theta)\).

Worked Examples

Why the Angular Condition Works

The center has polar direction \(\phi\), so a ray at angle \(\theta\) makes angle \(\theta-\phi\) with the line from the origin to the center. The perpendicular distance from the center to that line is \(d|\sin(\theta-\phi)|\). A line can meet a circle of radius \(R\) only if this distance is at most \(R\), which gives the discriminant condition \(d|\sin(\theta-\phi)|\le R\). When the origin is outside the disk, this line-distance test includes lines whose intersection lies behind the origin. Requiring \(q(\theta)>0\) selects the forward rays. This explains why both the discriminant and the sign of the radial roots matter.

The same fact follows algebraically. Completing the square in the radial quadratic gives

For the quadratic to be nonpositive, its squared term must be no greater than the remaining quantity. That quantity must therefore be nonnegative, producing the discriminant condition; the resulting interval must then be intersected with \(r\ge0\). This proves the radial-bound procedure directly from the circle inequality.

A Common Mistake: Keeping Only One Root

A frequent error is to treat an off-center circle as if its polar equation were a single radial function. For \((x-2)^2+y^2=1\), one might write \(r=2\cos\theta\), losing both the \(r^2\) term and the constant \(3\) in the correctly substituted equation \(r^2-4r\cos\theta+3=0\). At \(\theta=0\), the mistaken equation predicts the boundary radius \(r=2\). But substituting the point \((2,0)\) into the original circle equation gives \((2-2)^2=0\ne1\): it is inside, not on the boundary. The correct boundary radii at \(\theta=0\) are the two roots \(r=1\) and \(r=3\). A ray through an exterior disk enters and exits it, so both roots are needed.

Another important distinction is between the closed disk and its interior. The symbol \(\le\) includes the boundary; the strict inequality \(<\) describes only interior points. Polar bounds for a double integral over a disk are commonly written with inclusive limits because boundary points do not change the region’s area, but that convention does not make the boundary part of the interior.