question_answer

If the curve \[{{C}_{1}}\] is \[|z-1|\,=1\] and the point \[{{z}_{1}}\] satisfies the relation \[|{{z}_{1}}-8|+|{{z}_{1}}-6i=10,\] then the minimum value of \[|z-{{z}_{1}}|\] is

A)  \[\frac{16}{5}\]                                 

B)  \[\frac{\sqrt{10025}}{25}-1\]

C)  \[\frac{\sqrt{10125}}{25}-1\]                       

D)  24

Correct Answer: A

Solution :

 \[{{z}_{1}}\] lies on the line segemtn joining the point 8 and 6i, equation of the line is \[\frac{x}{8}+\frac{y}{6}=1\].                     ?(i) Slope is            \[-\frac{6}{8}=-\frac{3}{4}\]. A line is perpendicular to the line \[\frac{x}{8}+\frac{y}{4}=1\] and passing through the point \[(1,\,0)\] is \[y-0=\frac{4}{3}\,(x-1)\] \[\Rightarrow \]            \[4x-3y=4\] Point of intersection of lines (i) and (ii) is \[\left( \frac{88}{25},\,\,\frac{84}{25} \right)\] Min value of \[|z-{{z}_{1}}|\] is \[=\,\sqrt{{{\left( 1-\frac{88}{25} \right)}^{2}}+{{\left( 0-\frac{84}{25} \right)}^{2}}}-1\] \[=\sqrt{{{\left( \frac{63}{25} \right)}^{2}}\times \,{{\left( \frac{84}{25} \right)}^{2}}}-1\] \[=\frac{\sqrt{11025}}{25}-1=\frac{105}{25}-1\] \[=\frac{80}{25}=\frac{16}{5}\]