InterviewSolution
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InterviewSolution
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If `z=(lambda+3)+isqrt(5-lambda^2)` then the locus of Z is |
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Answer» `z = (lambda+3) +isqrt(5-lambda^2)` `=>x+iy = (lambda+3) +isqrt(5-lambda^2)` `=> x = lambda+3=> lambda = x-3` `=>lambda^2 = (x-3)^2->(1)` `=>y = sqrt(5-lambda^2) => y^2 = (5-lambda^2)` `=>lambda^2 = 5-y^2->(2)` From (1) and (2), `=> (x-3)^2 = 5-y^2` `=>(x-3)^2 +y^2 = 5` This is an equation of a circle with center `(3,0)` and radius `sqrt5`. So, option -`(b)` is the correct option. |
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