Solution 4.1:5a - Revision history

Tek at 10:47, 7 October 2008

2008-10-07T10:47:19Z

←Older revision		Revision as of 10:47, 7 October 2008
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-	~~{{NAVCONTENT_START}}~~	+	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
-	~~[[Image:4_1_5_a~~.~~gif\|center]]~~	+

-	{{~~NAVCONTENT_STOP~~}}	+	{{Displayed math\|\|<math>\sqrt{(x-1)^2 + (y-2)^2} = 2\,\textrm{.}</math>}}

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

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


-	~~<math>\sqrt{\left( x-1 \right)^{2}+\left( y-2 \right)^{2}}=2</math>~~	+	[[Image:4_1_5_a.gif\|center]]
-		+
-		+
-	~~After squaring, we obtain the equation of the circle in standard form~~:	+
-		+
-		+
-	~~<math>\left( x-1 \right)^{2}+\left( y-2 \right)^{2}=4</math>~~	+

2008-09-27T11:09:51Z

←Older revision		Revision as of 11:09, 27 September 2008
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	[[Image:4_1_5_a.gif\|center]]		[[Image:4_1_5_a.gif\|center]]

-	<center> [[Image:4_1_5a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+
		+	A circle is defined as all the points which have a fixed distance to the circle's midpoint. Hence, a point
		+	<math>\left( x \right.,\left. y \right)</math>
		+	lies on our circle if and only if its distance to the point
		+	<math>\left( 1 \right.,\left. 3 \right)</math>
		+	is exactly
		+	<math>2</math>. Using the distance formula, we can express this condition as
		+
		+
		+	<math>\sqrt{\left( x-1 \right)^{2}+\left( y-2 \right)^{2}}=2</math>
		+
		+
		+	After squaring, we obtain the equation of the circle in standard form:
		+
		+
		+	<math>\left( x-1 \right)^{2}+\left( y-2 \right)^{2}=4</math>

2008-09-09T09:36:03Z

←Older revision

Revision as of 09:36, 9 September 2008

2008-08-21T07:05:09Z

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2008-08-15T14:06:47Z

2008-04-02T13:01:52Z

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