A) \[\sqrt{13}\]
B) \[2\sqrt{13}\]
C) \[8\]
D) \[4+\sqrt{13}\]
Correct Answer: B
Solution :
Given equation represents the circle with centre \[(3,-2)\] and is of radius \[(R)=4\] \[|z|\] represents the distance of point 'z' from origin Greatest and least distances occur along the normal through the origin Normal always passes through center of circle From figure; let PQ be the normal through origin 'O' and C be its center \[(3,-2)\] it is clear that OP is the least distance and OQ is the greatest distance From diagram; \[OP=CP-OC\] and \[OQ=CQ+OC\] Here, \[CP=CQ=R=4\] \[OC=\sqrt{{{(3-0)}^{2}}+{{(-2-0)}^{2}}}\] \[\Rightarrow OC=\sqrt{13}\] \[\therefore OP=CP-OC\] \[\Rightarrow OP=4-\sqrt{13}\] \[\therefore \] Least distance \[OP=4-\sqrt{13}\] and \[OQ=CQ+OC\] \[\Rightarrow OQ=4+\sqrt{13}\] \[\therefore \]Greatest distance \[=OQ=4+\sqrt{13}\] Difference between greatest and least distance\[=OQ-OP=\left( 4+\sqrt{13} \right)-\left( 4-\sqrt{13} \right)\] \[\Rightarrow Difference=2\sqrt{13}\] final answer\[=2\sqrt{13}\] the correct option is 'B'You need to login to perform this action.
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