Solution 4.1:5b - Revision history

Tek at 10:53, 7 October 2008

Tek — Tue, 07 Oct 2008 10:53:17 GMT

←Older revision		Revision as of 10:53, 7 October 2008
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-	If the circle is to contain the point	+	If the circle is to contain the point (-1,1), then that point's distance away from the centre (2,-1) must equal the circle's radius, ''r''. Thus, we can obtain the circle's radius by calculating the distance between (-1,1) and (2,-1) using the distance formula,
-	~~<math>\left~~( -1 ~~\right.~~,~~\left.~~ 1 ~~\right~~)~~</math>~~, then that point's distance away from the centre	+
-	~~<math>\left~~( 2 ~~\right.~~,~~\left.~~ -1 ~~\right~~)~~</math>~~	+
-	must equal the circle's radius,	+
-	~~<math>~~r~~</math>~~. Thus, we can obtain the circle's radius by calculating the distance between	+
-	~~<math>\left~~( -1 ~~\right.~~,~~\left.~~ 1 ~~\right~~)~~</math>~~	+
-	and	+
-	~~<math>\left~~( 2 ~~\right.~~,~~\left.~~ -1 ~~\right~~)~~</math>~~	+
-	using the distance formula:	+
-		+
-		+
-	~~<math>\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}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	r &= \sqrt{(2-(-1))^2+(-1-1)^2} = \sqrt{3^2+(-2)^2} = \sqrt{9+4} = \sqrt{13}\,\textrm{.}
		+	\end{align}</math>}}

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

-		+	{{Displayed math\|\|<math>(x-2)^2 + (y-(-1))^2 = (\sqrt{13})^{2}</math>}}
-	<math>~~\left~~( x-2 ~~\right~~)^{2}+~~\left~~( y-~~\left~~( -1 ~~\right~~) ~~\right~~)^{2}=~~\left~~( \sqrt{13} ~~\right~~)^{2}</math>	+
-		+

	which the same as		which the same as

		+	{{Displayed math\|\|<math>(x-2)^{2} + (y+1)^2 = 13\,\textrm{.}</math>}}

-	<math>\left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13</math>

-	{{NAVCONTENT_START}}
	[[Image:4_1_5_b-1(2).gif\|center]]		[[Image:4_1_5_b-1(2).gif\|center]]

-	{{NAVCONTENT_STOP}}
-
-	NOTE: A circle having its centre at
-	<math>\left( a \right.,\left. b \right)</math>
-	and radius
-	<math>r</math>
-	has the equation

		+	Note: A circle having its centre at (''a'',''b'') and radius ''r'' has the equation

-	<math>~~\left~~( x-a ~~\right~~)^{2}+~~\left~~( y-b ~~\right~~)2=r^{2}</math>	+	{{Displayed math\|\|<math>(x-a)^2 + (y-b)^2 = r^2\,\textrm{.}</math>}}

Ian at 11:20, 27 September 2008

Ian — Sat, 27 Sep 2008 11:20:33 GMT

←Older revision		Revision as of 11:20, 27 September 2008
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		+	If the circle is to contain the point
		+	<math>\left( -1 \right.,\left. 1 \right)</math>, then that point's distance away from the centre
		+	<math>\left( 2 \right.,\left. -1 \right)</math>
		+	must equal the circle's radius,
		+	<math>r</math>. Thus, we can obtain the circle's radius by calculating the distance between
		+	<math>\left( -1 \right.,\left. 1 \right)</math>
		+	and
		+	<math>\left( 2 \right.,\left. -1 \right)</math>
		+	using the distance formula:
		+
		+
		+	<math>\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}</math>
		+
		+
		+	When we know the circle's centre and its radius, we can write the equation of the circle,
		+
		+
		+	<math>\left( x-2 \right)^{2}+\left( y-\left( -1 \right) \right)^{2}=\left( \sqrt{13} \right)^{2}</math>
		+
		+
		+	which the same as
		+
		+
		+	<math>\left( x-2 \right)^{2}+\left( y+1 \right)^{2}=13</math>
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	[[Image:4_1_5_b-1(2).gif\|center]]		[[Image:4_1_5_b-1(2).gif\|center]]

-	<center> [[Image:4_1_5b-1(2).gif]] </center>
-	{{NAVCONTENT_STOP}}
-	{{NAVCONTENT_START}}
-	<center> [[Image:4_1_5b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	NOTE: A circle having its centre at
		+	<math>\left( a \right.,\left. b \right)</math>
		+	and radius
		+	<math>r</math>
		+	has the equation
		+
		+
		+	<math>\left( x-a \right)^{2}+\left( y-b \right)2=r^{2}</math>

Tekbot: Lösning 4.1:5b moved to Solution 4.1:5b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:36:23 GMT

Lösning 4.1:5b moved to Solution 4.1:5b: Robot: moved page

←Older revision

Revision as of 09:36, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:05:19 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Martin at 14:11, 15 August 2008

Martin — Fri, 15 Aug 2008 14:11:55 GMT

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Oskar — Wed, 02 Apr 2008 13:02:43 GMT

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