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Te vjeter 26-12-2007, 20:17
Minifotoja e anetarit 8u1x0
8u1x0 8u1x0 eshte Jashte Linje
I/e Sapoardhur
 
Reg: 26-12-07
Postime: 1
Kryesore

Clearly the equation is of the form (x+a)^2 + (y+a)^2=a^2
Additionally we know that

(-4+a)^2+(2+a)^2 = 5^2+a^2
I get this from the fact that there must be a right triangle joining center of the circle and (-4,2)
or
16+a^2-8a+4+a^2+4a=25+a^2
-5+a^2-4a=0
a=(4+/-sqrt(16+20))/2
a is positive so = 5 (sqrt(36)+4)/2
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