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Solution 4.1:7c

From Förberedande kurs i matematik 1

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m (Lösning 4.1:7c moved to Solution 4.1:7c: Robot: moved page)
Current revision (11:31, 8 October 2008) (edit) (undo)
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By completing the square, we can rewrite the ''x''- and ''y''-terms as quadratic expressions,
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<center> [[Image:4_1_7_c.gif]] </center>
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{{Displayed math||<math>\begin{align}
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x^2 - 2x &= (x-1)^2 - 1^2\,,\\[5pt]
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y^2 + 6y &= (y+3)^2 - 3^2\,,
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\end{align}</math>}}
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<center> [[Image:4_1_7c.gif]] </center>
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and the whole equation then has standard form,
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{{NAVCONTENT_STOP}}
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{{Displayed math||<math>\begin{align}
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(x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt]
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(x-1)^2 + (y+3)^2 &= 7\,\textrm{.}
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\end{align}</math>}}
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From this, we see that the circle has its centre at (1,-3) and radius <math>\sqrt{7}\,</math>.
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<center> [[Image:4_1_7_c.gif]] </center>

Current revision

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

x2−2xy2+6y=(x−1)2−12=(y+3)2−32

and the whole equation then has standard form,

(x−1)2−1+(y+3)2−9(x−1)2+(y+3)2=−3=7.

From this, we see that the circle has its centre at (1,-3) and radius 7 .


Image:4_1_7_c.gif
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