// Numbas version: finer_feedback_settings {"name": "2.6.9 Circle proof", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "2.6.9 Circle proof", "tags": [], "metadata": {"description": "

Fixed question: Given two points at opposite ends of a diameter, write down the equation of the circle.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Show that if $A(x_1, y_1)$ and $B(x_2, y_2)$ are at opposite ends of a diameter of a circle then the equation of the circle is $(x − x_1)(x − x_2) + (y − y_1)(y − y_2) = 0$.

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Hint: if $P$ is any point on the circle, obtain the slopes of the lines $AP$ and $BP$ and recall that the angle in a semicircle must be a right angle.

", "advice": "

Let $P (x,y)$ be an arbitrary point on the circle.

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The the angle $\\angle APB$ is a right angle, meaning that $AP$ and $BP$ are perpendicular, meaning that the product of their gradients is $-1$.

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Let $m_1$ be the gradient of $AP$ and let $m_2$ be the gradient of $BP$.

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Then

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$m_1 = \\dfrac{y-y_1}{x-x_1}\\quad$ and $\\quad m_2=\\dfrac{y-y_2}{x-x_2}$

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So

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$m_1 m_2 = -1$

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$ \\dfrac{y-y_1}{x-x_1} \\times \\dfrac{y-y_2}{x-x_2}=-1$

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Multiplying through by the denominator:

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$(y-y_1)(y-y_2) = -(x-x_1)(x-x_2)$

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i.e.

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$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

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