Solution 4.1:5b

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m (Lösning 4.1:5b moved to Solution 4.1:5b: Robot: moved page)
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If the circle is to contain the point
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<math>\left( -1 \right.,\left. 1 \right)</math>, then that point's distance away from the centre
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<math>\left( 2 \right.,\left. -1 \right)</math>
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must equal the circle's radius,
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<math>r</math>. Thus, we can obtain the circle's radius by calculating the distance between
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<math>\left( -1 \right.,\left. 1 \right)</math>
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and
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<math>\left( 2 \right.,\left. -1 \right)</math>
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using the distance formula:
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<math>\begin{align}
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& r=\sqrt{\left( 2-\left( -1 \right) \right)^{2}+\left( -1-1 \right)^{2}}=\sqrt{3^{2}+\left( -2 \right)^{2}} \\
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& =\sqrt{9+4}=\sqrt{13} \\
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\end{align}</math>
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When we know the circle's centre and its radius, we can write the equation of the circle,
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<math>\left( x-2 \right)^{2}+\left( y-\left( -1 \right) \right)^{2}=\left( \sqrt{13} \right)^{2}</math>
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which the same as
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<math>\left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13</math>
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[[Image:4_1_5_b-1(2).gif|center]]
[[Image:4_1_5_b-1(2).gif|center]]
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<center> [[Image:4_1_5b-1(2).gif]] </center>
 
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<center> [[Image:4_1_5b-2(2).gif]] </center>
 
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NOTE: A circle having its centre at
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<math>\left( a \right.,\left. b \right)</math>
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and radius
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<math>r</math>
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has the equation
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 +
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<math>\left( x-a \right)^{2}+\left( y-b \right)2=r^{2}</math>

Revision as of 11:20, 27 September 2008

If the circle is to contain the point \displaystyle \left( -1 \right.,\left. 1 \right), then that point's distance away from the centre \displaystyle \left( 2 \right.,\left. -1 \right) must equal the circle's radius, \displaystyle r. Thus, we can obtain the circle's radius by calculating the distance between \displaystyle \left( -1 \right.,\left. 1 \right) and \displaystyle \left( 2 \right.,\left. -1 \right) using the distance formula:


\displaystyle \begin{align} & r=\sqrt{\left( 2-\left( -1 \right) \right)^{2}+\left( -1-1 \right)^{2}}=\sqrt{3^{2}+\left( -2 \right)^{2}} \\ & =\sqrt{9+4}=\sqrt{13} \\ \end{align}


When we know the circle's centre and its radius, we can write the equation of the circle,


\displaystyle \left( x-2 \right)^{2}+\left( y-\left( -1 \right) \right)^{2}=\left( \sqrt{13} \right)^{2}


which the same as


\displaystyle \left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13

NOTE: A circle having its centre at \displaystyle \left( a \right.,\left. b \right) and radius \displaystyle r has the equation


\displaystyle \left( x-a \right)^{2}+\left( y-b \right)2=r^{2}

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