A) 7
B) 10
C) 12
D) None of these
Correct Answer: C
Solution :
There are 4 such circles, two of which have centres in first quadrant (h, h) and other two have centres in second (-h, h) and fourth quadrant (h, -h) respectively. \[\left| \frac{4h+3h-12}{5} \right|=h\,\Rightarrow \,h=6,\,1\] Radii in first quadrant \[-\frac{-4h+3h-12}{5}=h\Rightarrow \,h=3\] Radius in second quadrant. \[-\frac{4h-3h-12}{5}=h\Rightarrow h=2\] Radius in second quadrant. Hence sum = 12 Alternatively: \[\frac{\Delta }{s}+\frac{\Delta }{s-a}+\frac{\Delta }{s-b}+\frac{\Delta }{s-c}=\frac{6}{6}+\frac{6}{6-3}+\frac{6}{6-4}+\frac{6}{6-5}\]\[=1+2+3+6=12\]You need to login to perform this action.
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