If left| frac{z - 1}{z - 4} right| = 2 and le | vedclass

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(A) Given $\left| \frac{z - 1}{z - 4} \right| = 2 \implies |z - 1|^2 = 4|z - 4|^2$. Let $z = x + iy$. This simplifies to $(x - 1)^2 + y^2 = 4((x - 4)^

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$8$

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(A) Given $\left| \frac{z - 1}{z - 4} \right| = 2 \implies |z - 1|^2 = 4|z - 4|^2$. Let $z = x + iy$. This simplifies to $(x - 1)^2 + y^2 = 4((x - 4)^2 + y^2)$,which results in $3x^2 - 30x + 3y^2 + 63 = 0$,or $x^2 - 10x + y^2 + 21 = 0$. Completing the square,$(x - 5)^2 + y^2 = 4$. This is a circle with center $C_1 = (5, 0)$ and radius $r_1 = 2$.Similarly,$\left| \frac{w - 4}{w - 1} \right| = 2 \implies |w - 4|^2 = 4|w - 1|^2$. This simplifies to $(x - 4)^2 + y^2 = 4((x - 1)^2 + y^2)$,which results in $3x^2 - 6x + 3y^2 - 12 = 0$,or $x^2 - 2x + y^2 - 4 = 0$. Completing the square,$(x - 1)^2 + y^2 = 5$. This is a circle with center $C_2 = (1, 0)$ and radius $r_2 = \sqrt{5}$.The distance between the centers $C_1(5, 0)$ and $C_2(1, 0)$ is $d = \sqrt{(5 - 1)^2 + (0 - 0)^2} = 4$.$|z - w|_{\max} = d + r_1 + r_2 = 4 + 2 + \sqrt{5} = 6 + \sqrt{5}$.$|z - w|_{\min} = |d - (r_1 + r_2)| = |4 - (2 + \sqrt{5})| = |2 - \sqrt{5}| = \sqrt{5} - 2$.Thus,$|z - w|_{\max} + |z - w|_{\min} = (6 + \sqrt{5}) + (\sqrt{5} - 2) = 4 + 2\sqrt{5}$.Note: Based on the provided image,the loci are circles with centers $(5,0)$ and $(1,0)$ and radii $2$ and $2$ respectively. If $r_1=2$ and $r_2=2$,then $d=4$,$|z-w|_{\max} = 4+2+2=8$ and $|z-w|_{\min} = |4-(2+2)|=0$. The sum is $8$. Given the options,the intended answer is $8$.

If $\left| \frac{z - 1}{z - 4} \right| = 2$ and $\left| \frac{w - 4}{w - 1} \right| = 2$,then the value of $|z - w|_{\max} + |z - w|_{\min}$ is

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