// Numbas version: exam_results_page_options {"name": "Graphing: semi-circles centred at the origin", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variables": {"r": {"definition": "shuffle(1..12)[0..5]", "templateType": "anything", "group": "Ungrouped variables", "name": "r", "description": ""}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Recognising the equation of a semi-circle centred at the origin. Top, bottom, left and right.

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Being able to identify the equation of a semi-circle can come in handy, not just as an exercise in graphing, relations or functions, but also for more advanced things like integration using trigonometric substitution. 

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The equation of a circle centred at the origin of radius $r$ is normally written as \\[x^2+y^2=r^2.\\] 

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Solving for $y$ gives the equation \\[y=\\pm\\sqrt{r^2-x^2}.\\]

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This is the equation for the same circle, it is just written differently. Notice we needed to take the plus or minus square-root to undo the square. As you would expect, the plus sign allows $y$ to have positive values and the minus sign allows $y$ to have positive values. This means the equation \\[y=\\sqrt{r^2-x^2}\\] is the top half of the circle centred at the origin with radius $r$, and \\[y=-\\sqrt{r^2-x^2}\\] is the bottom half of the circle.

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Solving the original equation of the circle for $x$ gives the equation \\[x=\\pm\\sqrt{r^2-y^2}.\\]

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Again, this is the equation for the same circle, it is just written differently. Notice the plus or minus sign giving both positive and negative $x$ values. If we only want the right hand side of the circle we restrict ourselves to the positive $x$ values and so the equation of that semi-circle would be \\[x=\\sqrt{r^2-y^2}.\\]

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However, if we only want the left hand side of the circle we restrict ourselves to the negative $x$ values, and our equation would be \\[x=-\\sqrt{r^2-y^2}.\\]

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The left half of a circle

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The right half of a circle

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The top half of a circle

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The bottom half of a circle

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An entire circle

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Match the equations with the descriptions by clicking on the corresponding buttons.

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$x=-\\sqrt{\\var{r[0]}^2-y^2}$

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$x=\\sqrt{\\var{r[1]}^2-y^2}$

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$y=\\sqrt{\\var{r[2]}^2-x^2}$

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$y=-\\sqrt{\\var{r[3]}^2-x^2}$

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$y=\\pm\\sqrt{\\var{r[4]}^2-x^2}$

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