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Solution 4.1:7a

From Förberedande kurs i matematik 1

Revision as of 11:52, 27 September 2008 by Ian (Talk | contribs)
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As the equation stands, it is difficult directly to know anything about the circle, but if we complete the square and combine x - and y - terms together in their own respective square terms, then we will have the equation in the standard form,


xa2+yb2=r2 


and we will then be able to read off the circle's centre and radius.

If we take the x - and y - terms on the left-hand side and complete the square, we get


x2+2x=x+1212 


y22y=y1212 

and then the whole equation can be written as


x+1212+y1212=1 


or, with the constants moved to the right-hand side,


x+12+y12=3 


This is a circle having its centre at 11  and radius 3 .