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Analyse semicircles and circular regions

Start a set of 8
Step 1 of 5

Which first step selects the upper semicircle from the circle equation?

\[\left(x-2\right)^2+\left(y+1\right)^2=25\]
Year 11, VCE Units 1 & 2, Chapter 6, 4 difficulty levels, No calculatorBuild a worksheet

Semicircles and circular regions

A circle equation can be rearranged into an upper or lower square-root branch. The radicand gives the domain, and squared-distance comparisons describe circular regions without needing a square root

\[y=k\pm\sqrt{r^2-\left(x-h\right)^2}\]
  1. Read the centre and radius The centre coordinates have opposite signs to the signs inside the brackets. The radius is the positive square root of the squared radius \(\displaystyle \left(x-2\right)^2+\left(y+1\right)^2=25\Rightarrow C=(2,-1),\quad r=5\)
  2. Choose the named branch Use the positive square root for the upper branch and the negative square root for the lower branch \(\displaystyle y=-1+\sqrt{25-\left(x-2\right)^2}\text{ is the upper branch}\)
  3. Use the radicand for the domain The square-root radicand must be non-negative, and equality includes the two endpoints \(\displaystyle \left(x-2\right)^2\le25\Rightarrow -3\le x\le7\)
  4. Compare squared distances Use less than, equal to or greater than the squared radius to classify a point or select an inside or outside region \(\displaystyle \left(x-2\right)^2+\left(y+1\right)^2\le25\text{ includes the circular boundary}\)

Worked example

For the lower semicircle of the circle shown below, state its rule, domain and range

\[\left(x+1\right)^2+\left(y-3\right)^2=36\]
  1. \(\displaystyle \left(y-3\right)^2=36-\left(x+1\right)^2\)Isolate the squared vertical difference from the circle equation
  2. \(\displaystyle y=3-\sqrt{36-\left(x+1\right)^2}\)The lower semicircle is below the centre, so choose the negative square root
  3. \(\displaystyle \left(x+1\right)^2\le36\Rightarrow -7\le x\le5\)The radicand is non-negative, giving the closed domain
  4. \(\displaystyle \operatorname{Dom}=[-7,5],\quad\operatorname{Ran}=[-3,3]\)The lowest point is (-1,-3), and the endpoints are (-7,3) and (5,3)

Common mistakes

Choosing the wrong square-root branch
✕\(\displaystyle \left(x-1\right)^2+\left(y-2\right)^2=16,\quad y=2+\sqrt{16-\left(x-1\right)^2}\)✓\(\displaystyle \left(x-1\right)^2+\left(y-2\right)^2=16,\quad y=2-\sqrt{16-\left(x-1\right)^2}\)
A lower semicircle uses the negative square-root branch on or below the centre
Centre signs are copied
✕\(\displaystyle \left(x-4\right)^2+\left(y+2\right)^2=25\Rightarrow C=(-4,2)\)✓\(\displaystyle \left(x-4\right)^2+\left(y+2\right)^2=25\Rightarrow C=(4,-2)\)
The centre coordinates have opposite signs to the signs inside the brackets
The circular boundary is made open
✕\(\displaystyle \left(x-2\right)^2+\left(y+1\right)^2<25\)✓\(\displaystyle \left(x-2\right)^2+\left(y+1\right)^2\le25\)
A non-strict inequality includes points on the circular boundary

Difficulty levels

introSelect a branch and its range for a base semicircle, circular region or point
standardAnalyse a translated branch and state its exact domain and range
challengeHandle strict or non-strict circular regions and recover a parent circle from a branch
examSelect a regenerated formula or region statement with branch, centre and boundary traps

Questions about this topic

Why does the upper semicircle use the positive square root?+

The upper branch has y-values at or above the centre y-coordinate, so the positive square-root term is used

Why are the domain endpoints included?+

At each endpoint the radicand is zero, which gives a real point on the circle

How can a point be classified without finding a distance?+

Calculate its squared distance from the centre and compare it directly with the squared radius

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 6: Functions and relations

In this chapter

Classify a relation and state its domain and rangeEvaluate and determine function rulesFind the maximal domain of an algebraic functionAnalyse reciprocal and square-reciprocal functionsAnalyse sideways-parabola and root-function familiesDetermine a power-family rule from features and pointsDetermine a circle's equation and featuresAnalyse semicircles and circular regionsApply and describe transformations of a functionIntersections and tangents of a line and a circleEvaluate and interpret a restricted rational modelForm and analyse an inverse relation or function