From Förberedande kurs i matematik 1

Revision as of 09:38, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.1:7c moved to Solution 4.1:7c: Robot: moved page) ← Previous diff		Revision as of 12:06, 27 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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		+	By completing the square, we can rewrite the
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	-terms as quadratic expressions,
		+
		+
		+	<math>x^{2}-2x=\left( x-1 \right)^{2}-1^{2}</math>
		+
		+
		+	<math>y^{2}+6y=\left( y+3 \right)^{2}-3^{2}</math>
		+
		+	and the whole equation then has standard form,
		+
		+
		+	<math>\begin{align}
		+	& \left( x-1 \right)^{2}-1+\left( y+3 \right)^{2}-9=-3 \\
		+	& \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+3 \right)^{2}=7 \\
		+	\end{align}</math>
		+
		+
		+	From this, we see that the circle has its centre at
		+	<math>\left( 1 \right.,\left. -3 \right)</math>
		+	and radius
		+	<math>\sqrt{7}</math>.
		+
		+
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	<center> [[Image:4_1_7_c.gif]] </center>		<center> [[Image:4_1_7_c.gif]] </center>
-	<center> [[Image:4_1_7c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

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Revision as of 12:06, 27 September 2008

By completing the square, we can rewrite the \displaystyle x - and \displaystyle y -terms as quadratic expressions,

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

\displaystyle y^{2}+6y=\left( y+3 \right)^{2}-3^{2}

and the whole equation then has standard form,

\displaystyle \begin{align} & \left( x-1 \right)^{2}-1+\left( y+3 \right)^{2}-9=-3 \\ & \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+3 \right)^{2}=7 \\ \end{align}

From this, we see that the circle has its centre at \displaystyle \left( 1 \right.,\left. -3 \right) and radius \displaystyle \sqrt{7}.