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Recognising the equation of a semi-circle centred at the origin. Top, bottom, left and right.
"}, "statement": "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.
", "name": "Graphing: semi-circles centred at the origin", "variablesTest": {"maxRuns": 100, "condition": ""}, "extensions": [], "rulesets": {}, "tags": ["circles", "Graphing", "graphing", "semi-circles"], "advice": "The equation of a circle centred at the origin of radius $r$ is normally written as \\[x^2+y^2=r^2.\\]
\nSolving for $y$ gives the equation \\[y=\\pm\\sqrt{r^2-x^2}.\\]
\nThis is the equation for the same circle, it is just written differently. Notice we needed to take the plus or minus
Solving the original equation of the circle for $x$ gives the equation \\[x=\\pm\\sqrt{r^2-y^2}.\\]
\nAgain, 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
However, if we only want the
The left half of a circle
", "The right half of a circle
", "The top half of a circle
", "The bottom half of a circle
", "An entire circle
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", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "showFeedbackIcon": true, "displayType": "radiogroup", "answers": ["$x=-\\sqrt{\\var{r[0]}^2-y^2}$
", "$x=\\sqrt{\\var{r[1]}^2-y^2}$
", "$y=\\sqrt{\\var{r[2]}^2-x^2}$
", "$y=-\\sqrt{\\var{r[3]}^2-x^2}$
", "$y=\\pm\\sqrt{\\var{r[4]}^2-x^2}$
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