A) 6
B) \[6\pm 8i\]
C) \[6+8i,\,6+17i\]
D) None of these
Correct Answer: C
Solution :
We have \[\left| \frac{z-12}{z-8i} \right|=\frac{5}{3}\]and \[\left| \frac{z-4}{z-8} \right|=1\] Let\[z=x+iy\], then \[\left| \frac{z-12}{z-8i} \right|=\frac{5}{3}\Rightarrow 3|z-12|=5|z-8i|\] Þ \[3|(x-12)+iy|=5|x+(y-8)i|\] Þ \[9{{(x-12)}^{2}}+9{{y}^{2}}=25{{x}^{2}}+25{{(y-8)}^{2}}\] ....(i) and \[\left| \frac{z-4}{z-8} \right|=1\Rightarrow |z-4|=|z-8|\] Þ \[|x-4+iy|=|x-8+iy|\] Þ \[{{(x-4)}^{2}}+{{y}^{2}}={{(x-8)}^{2}}+{{y}^{2}}\Rightarrow x=6\] Putting \[x=6\]in (i), we get \[{{y}^{2}}-25y+136=0\] \ \[y=17,8\] Hence \[z=6+17i\]or \[z=6+8i\] Trick: Check it with options.
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