Bounds: Circle Circumference & Area

Statement

When the radius of a circle is only known within bounds, we can push those bounds through the formulas for circumference and area because both formulas increase as the radius increases. If the radius lies in the interval \(r \in [L_r,\,U_r]\) with \(0 \le L_r \le U_r\), then the circumference \(C=2\pi r\) and area \(A=\pi r^2\) satisfy

\[ C \in [\,2\pi L_r,\; 2\pi U_r\,], \qquad A \in [\,\pi L_r^{2},\; \pi U_r^{2}\,]. \]

Here \(L_r\) and \(U_r\) denote the lower and upper bounds of the radius. We usually leave answers exact in \(\pi\) unless a decimal is requested.

Why it’s true (short reason)

• For \(r\ge 0\), the functions \(f(r)=2\pi r\) and \(g(r)=\pi r^2\) are increasing (monotonic). Increasing functions send smaller inputs to smaller outputs and larger inputs to larger outputs.
• Therefore, the smallest possible circumference/area occurs when \(r=L_r\), and the largest occurs when \(r=U_r\). Substituting the endpoints gives the output interval.
• Because area depends on \(r^2\), a small percentage change in \(r\) produces roughly double the percentage change in \(A\). This is why area bounds are “wider” (in relative terms) than circumference bounds.

Recipe (how to use it)

1. Build the radius interval. If the radius was rounded, convert the rounding statement to bounds:
   • Nearest cm: \(x\) cm \(\Rightarrow r \in [x-0.5,\,x+0.5)\) cm.
   • To \(d\) decimal places: step \(=0.5\times 10^{-d}\). So \(x\) \(\Rightarrow [\,x-\text{step},\,x+\text{step})\).
   • To \(s\) significant figures: find the place value of the last significant digit and take half that as the step.
2. Circumference. Use \(C_{\min}=2\pi L_r\), \(C_{\max}=2\pi U_r\). Keep the unit the same as the radius (cm, m, km, …).
3. Area. Use \(A_{\min}=\pi L_r^{2}\), \(A_{\max}=\pi U_r^{2}\). Units are squared (cm\(^2\), m\(^2\), …).
4. Exact vs decimal. Unless told otherwise, leave bounds exact in \(\pi\) (e.g., \(23\pi\) cm). If a decimal is required, round only at the end to the stated accuracy.
5. Reverse questions. If a rounded circumference or area is given, first bound that quantity, then convert to radius using \(r=\dfrac{C}{2\pi}\) or \(r=\sqrt{A/\pi}\); finally, if needed, push the new radius bounds back through to get bounds for the other quantity.

Spotting it

Use this technique whenever you see:

• “Radius \(=\) … to the nearest … / to 2 d.p. / to 2 s.f.”
• “Find the maximum or minimum possible circumference/area.”
• “A circular object with measured radius/diameter; give bounds for \(C\) or \(A\).”
• “Circumference (or area) measured to a given accuracy; find bounds for the radius.”

Common pairings

• Rounding to bounds (constructing \([L_r,\,U_r]\) correctly).
• Percentage error (comparing relative uncertainty in \(C\) vs \(A\)).
• Unit conversions (mm–cm–m) before or after applying the formulas.
• Diameter information (use \(r=\tfrac{d}{2}\) and bound the diameter first).

Mini examples

1. Given: \(r\in[4.8,5.2]\) cm. Find: bounds for \(C\). Answer: \(C\in[2\pi\cdot 4.8,\,2\pi\cdot 5.2]=[9.6\pi,\,10.4\pi]\) cm.
2. Given: \(r=8.40\) cm (2 d.p.). \strong>Find: bounds for \(A\). Answer: \(r\in[8.395,8.405)\) so \(A\in[\pi\cdot 8.395^{2},\,\pi\cdot 8.405^{2})=[70.48\pi,\,70.64\pi)\) cm\(^2\).
3. Given: \(C=50\) m (nearest m). Find: bounds for \(r\). Answer: \(C\in[49.5,50.5)\) so \(r\in\big[\tfrac{49.5}{2\pi},\,\tfrac{50.5}{2\pi}\big)\) m.

Working carefully with rounding

Most GCSE questions phrase measurements “to the nearest unit” or “to a number of decimal places”. Translate these into half-step intervals centred at the stated value, with an open upper end. For instance:

• \(12\) cm to nearest cm \(\Rightarrow [11.5,12.5)\) cm.
• \(0.45\) m to 2 s.f. \(\Rightarrow [0.445,0.455)\) m (because the last s.f. is the hundredths place).
• \(7.6\) m to 1 d.p. \(\Rightarrow [7.55,7.65)\) m.

Once \([L_r,\,U_r]\) is set, the rest is substitution. Keep results exact in \(\pi\) unless a decimal is requested.

Reverse problems (given \(C\) or \(A\), then radius)

Given circumference

1. Bound the circumference from the rounding (e.g., nearest metre: \([C-0.5,\,C+0.5)\)).
2. Convert to radius bounds: \(\displaystyle r\in\Big[\frac{C_{\min}}{2\pi},\,\frac{C_{\max}}{2\pi}\Big)\).
3. If needed, push to area: \(\displaystyle A\in\Big[\pi\Big(\frac{C_{\min}}{2\pi}\Big)^{2},\,\pi\Big(\frac{C_{\max}}{2\pi}\Big)^{2}\Big]=\Big[\frac{C_{\min}^{2}}{4\pi},\,\frac{C_{\max}^{2}}{4\pi}\Big)\).

Given area

1. Bound the area (e.g., nearest \(10\pi\) cm\(^2\) gives \([A-5\pi,\,A+5\pi)\)).
2. Convert to radius bounds using the increasing square root: \(\displaystyle r\in\Big[\sqrt{\frac{A_{\min}}{\pi}},\,\sqrt{\frac{A_{\max}}{\pi}}\Big)\).
3. If needed, push to circumference: \(C\in[\,2\pi L_r,\,2\pi U_r\,]\).

Percentage uncertainty and sensitivity

Suppose \(r\) is measured with small relative error \(\delta r/r\). Then

• For \(C=2\pi r\): \(\dfrac{\delta C}{C}\approx\dfrac{\delta r}{r}\). Relative uncertainty is the same as the radius’s.
• For \(A=\pi r^{2}\): \(\dfrac{\delta A}{A}\approx 2\,\dfrac{\delta r}{r}\). Relative uncertainty is roughly double.

Interpretation: area magnifies radius error more than circumference does. In exams, this helps you sanity-check whether your area bounds look proportionally wider than your circumference bounds.

Units and presentation

• Keep units with the final bounds: circumference uses the same length unit as radius; area uses squared units.
• When leaving answers exact in \(\pi\), place the unit after the expression: e.g., \(23\pi\) cm, \(70.48\pi\) cm\(^2\).
• When decimals are required, evaluate with \(\pi\) at the end and round to the stated accuracy.

Pitfalls

• Forgetting to form radius bounds first. Do not substitute a rounded centre value straight into the formulas; always create \([L_r,\,U_r]\) from the rounding statement.
• Upper endpoint inclusion. Rounded bounds typically use an open upper end, e.g., \([11.5,12.5)\). Keep that openness when you push through increasing functions.
• Mixing exact and decimal forms. Don’t write \(9.6\pi\) and then multiply by \(\pi\) again; choose either exact-in-\(\pi\) or a rounded decimal once.
• Unit slips. Circumference is in cm/m; area is in cm\(^2\)/m\(^2\). State units clearly for both lower and upper bounds.
• Using diameter by mistake. If given diameter bounds, convert to radius bounds via halving before applying the formulas.

Exam strategy

• Underline the rounding phrase (“nearest…”, “to 2 d.p.”) and immediately write the numeric interval for \(r\) (or for \(C\)/\(A\) if those are the rounded quantities).
• Work exact in \(\pi\) to avoid rounding drift; only round at the end if asked.
• State both ends clearly (e.g., \(C_{\min}=\dots\), \(C_{\max}=\dots\)), then present the interval in one line with units.
• Reverse carefully when given \(C\) or \(A\): bound the given quantity first, then convert to \(r\), not the other way round.
• Sanity-check orders of magnitude (e.g., if \(r\approx 5\) cm, then \(C\) should be around \(10\pi\) cm, \(A\) around \(25\pi\) cm\(^2\)).

Extended micro-examples (quick practice)

1. Linear sensitivity: \(r\in[3,3.1]\) km. Then \(C\in[6\pi,6.2\pi]\) km. The width grows in proportion to \(r\).
2. Area amplification: \(r\in[15,16]\) mm. Then \(A\in[\pi\cdot 15^{2},\,\pi\cdot 16^{2}]=[225\pi,256\pi]\) mm\(^2\). Relative spread is larger than for \(C\).
3. Two-stage reverse: \(C=94\) cm to nearest cm \(\Rightarrow C\in[93.5,94.5)\) cm \(\Rightarrow r\in\big[\tfrac{93.5}{2\pi},\,\tfrac{94.5}{2\pi}\big)\) cm. Then \(A\in\big[\tfrac{93.5^{2}}{4\pi},\,\tfrac{94.5^{2}}{4\pi}\big)\) cm\(^2\).
4. Significant figures: \(r=0.072\) m to 2 s.f. The last s.f. is the hundredths of a metre? No—here it is the second non-zero digit (the thousandths place overall). Bounds: \(r\in[0.0715,0.0725)\) m. Then \(C\in[2\pi\cdot 0.0715,\,2\pi\cdot 0.0725)\) m, \(A\in[\pi\cdot 0.0715^{2},\,\pi\cdot 0.0725^{2})\) m\(^2\).

Summary

To bound a circle’s circumference or area, first translate any rounding into a precise interval for the radius. Because \(2\pi r\) and \(\pi r^2\) are increasing for \(r\ge 0\), substitute the radius bounds directly to get tight bounds: \[ C \in [\,2\pi L_r,\; 2\pi U_r\,], \qquad A \in [\,\pi L_r^{2},\; \pi U_r^{2}\,]. \] Work exact in \(\pi\), attach correct units (length for \(C\), squared length for \(A\)), and only round to decimals at the very end if required. In reverse problems, bound \(C\) or \(A\) first, then convert to \(r\) using \(r=\dfrac{C}{2\pi}\) or \(r=\sqrt{A/\pi}\). Watch the open upper endpoint inherited from rounding, and remember that area is more sensitive than circumference to errors in the radius. With these habits, you can produce clear, fully justified bounds every time.