Lösung 4.1:5a

Aus Online Mathematik Brückenkurs 1

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A circle is defined as all the points which have a fixed distance to the circle's midpoint. Hence, a point
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<math>\left( x \right.,\left. y \right)</math>
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lies on our circle if and only if its distance to the point
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<math>\left( 1 \right.,\left. 3 \right)</math>
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is exactly
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<math>2</math>. Using the distance formula, we can express this condition as
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<math>\sqrt{\left( x-1 \right)^{2}+\left( y-2 \right)^{2}}=2</math>
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After squaring, we obtain the equation of the circle in standard form:
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<math>\left( x-1 \right)^{2}+\left( y-2 \right)^{2}=4</math>

Version vom 11:09, 27. Sep. 2008


A circle is defined as all the points which have a fixed distance to the circle's midpoint. Hence, a point \displaystyle \left( x \right.,\left. y \right) lies on our circle if and only if its distance to the point \displaystyle \left( 1 \right.,\left. 3 \right) is exactly \displaystyle 2. Using the distance formula, we can express this condition as


\displaystyle \sqrt{\left( x-1 \right)^{2}+\left( y-2 \right)^{2}}=2


After squaring, we obtain the equation of the circle in standard form:


\displaystyle \left( x-1 \right)^{2}+\left( y-2 \right)^{2}=4

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:5a“