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Solution 4.1:7c

From Förberedande kurs i matematik 1

Revision as of 11:31, 8 October 2008 by Tek (Talk | contribs)
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By completing the square, we can rewrite the x- and y-terms as quadratic expressions,

x22xy2+6y=(x1)212=(y+3)232

and the whole equation then has standard form,

(x1)21+(y+3)29(x1)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