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Solution 4.1:5a

From Förberedande kurs i matematik 1

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Current revision (10:47, 7 October 2008) (edit) (undo)
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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 (''x'',''y'') lies on our circle if and only if its distance to the point (1,3) is exactly 2. Using the distance formula, we can express this condition as
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[[Image:4_1_5_a.gif|center]]
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{{Displayed math||<math>\sqrt{(x-1)^2 + (y-2)^2} = 2\,\textrm{.}</math>}}
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After squaring, we obtain the equation of the circle in standard form,
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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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{{Displayed math||<math>(x-1)^2 + (y-2)^2 = 4\,\textrm{.}</math>}}
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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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[[Image:4_1_5_a.gif|center]]
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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>
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Current revision

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

(x−1)2+(y−2)2=2. 

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

(x−1)2+(y−2)2=4.


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