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Question Number 32049 by rahul 19 last updated on 18/Mar/18

If ∣z−6−8i∣≤4 then Minimum value  of ∣z∣ is   A) 4   B) 5  C) 6  D) 8.

$$\boldsymbol{{I}}{f}\:\mid\boldsymbol{{z}}−\mathrm{6}−\mathrm{8}\boldsymbol{{i}}\mid\leqslant\mathrm{4}\:{then}\:\boldsymbol{{M}}{inimum}\:{value} \\ $$$${of}\:\mid\boldsymbol{{z}}\mid\:{is}\: \\ $$$$\left.{A}\right)\:\mathrm{4}\: \\ $$$$\left.{B}\right)\:\mathrm{5} \\ $$$$\left.{C}\right)\:\mathrm{6} \\ $$$$\left.{D}\right)\:\mathrm{8}. \\ $$

Answered by mrW2 last updated on 18/Mar/18

(√(6^2 +8^2 ))−4=6  ⇒answer C)

$$\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} }−\mathrm{4}=\mathrm{6} \\ $$$$\left.\Rightarrow{answer}\:{C}\right) \\ $$

Commented by rahul 19 last updated on 18/Mar/18

thank u sir!

$${thank}\:{u}\:{sir}! \\ $$

Answered by Tinkutara last updated on 18/Mar/18

Commented by Tinkutara last updated on 18/Mar/18

In question ∣z−6−8i∣≤4  you wrote ∣z−6−8i∣≤∣z∣+∣−6−8i∣  In general a≤b and a≤c⇏b=c  e.g. 5≤10 and 5≤20 but 10≠20  We cannot compare b and c.

$${In}\:{question}\:\mid{z}−\mathrm{6}−\mathrm{8}{i}\mid\leqslant\mathrm{4} \\ $$$${you}\:{wrote}\:\mid{z}−\mathrm{6}−\mathrm{8}{i}\mid\leqslant\mid{z}\mid+\mid−\mathrm{6}−\mathrm{8}{i}\mid \\ $$$${In}\:{general}\:{a}\leqslant{b}\:{and}\:{a}\leqslant{c}\nRightarrow{b}={c} \\ $$$${e}.{g}.\:\mathrm{5}\leqslant\mathrm{10}\:{and}\:\mathrm{5}\leqslant\mathrm{20}\:{but}\:\mathrm{10}\neq\mathrm{20} \\ $$$${We}\:{cannot}\:{compare}\:{b}\:{and}\:{c}. \\ $$

Commented by rahul 19 last updated on 18/Mar/18

check my method , where am i wrong?

$${check}\:{my}\:{method}\:,\:{where}\:{am}\:{i}\:{wrong}? \\ $$

Commented by rahul 19 last updated on 18/Mar/18

ok,thanks!

$${ok},{thanks}! \\ $$

Commented by Tinkutara last updated on 18/Mar/18

Locus of z is a circle and its inside. Hence calculating the minimum distance from origin.

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