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STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
If $|z - 2|/|z - 3| = 2$ represents a circle, then its radius is equal to
  • A
    $1$
  • B
    $1/3$
  • C
    $3/4$
  • ✓
    $2/3$
✓

Answer

Correct option: D.
$2/3$
d
(d)Given, $\frac{{|z - 2|}}{{|z - 3|}} = 2$
==> $\sqrt {{{(x - 2)}^2} + {y^2}} = 2\sqrt {{{(x - 3)}^2} + {y^2}} $
==> ${(x - 2)^2} + {y^2} = 4[{(x - 3)^2} + {y^2}]$
==> ${x^2} + {y^2} + 4 - 4x = 4{x^2} + 4{y^2} + 36 - 24x$
==> $3{x^2} + 3{y^2} - 20x + 32 = 0$
or ${x^2} + {y^2} - \frac{{20}}{3}x + \frac{{32}}{3} = 0$ .....$(i)$
We know that, standard equation of circle,
${x^2} + {y^2} + 2gx + 2fy + c = 0$ .....$(ii)$
Comparison of $(i)$ from $(ii)$
==> $2g = - \frac{{20}}{3}\, \Rightarrow g = - \frac{{10}}{3},f = 0,c = \frac{{32}}{3}$
Hence, Radius =$\sqrt {{g^2} + {f^2} - c} $$ = \sqrt {\frac{{100}}{9} - \frac{{32}}{3}} = \sqrt {\frac{4}{9}} = \frac{2}{3}$

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