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The points of intersection of the two curves `|z-3|=2` and `|z|=2` in an argand plane are: |
Answer» `Z=x+iy` `|z|=sqrt(x^2+y^2)=2` `x^2+y^2=2` `Z-3=x+iy-3=(x-3)+iy` `|z-3|=sqrt((x-3)^2+y^2)=2` `(x-3)^2+y^2=4-(2)` `cx^2-(x-3)^2=0` `x^2=(x-3)^2` `x=pm(x-3)` `x=x-3` `x=-x+3` `2x=3` `x=3/2` `y^2=4-9/4` `y=sqrt7/2` Intersection=`3/2+sqrt7/2i` and `3/2-sqrt7/2i` `1/2(3pmisqrt7)` option 2 is corrrect. |
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