If {z_1} = 10 + 6i,{z_2} = 4 + 6i and z is a | vedclass

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(C) Given $z_1 = 10 + 6i$ and $z_2 = 4 + 6i$. Let $z = x + iy$.The condition $	ext{amp}\left( \frac{z - z_1}{z - z_2} ight) = \frac{\pi}{4}$ repres

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$3\sqrt{2}$

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(C) Given $z_1 = 10 + 6i$ and $z_2 = 4 + 6i$. Let $z = x + iy$.The condition $	ext{amp}\left( \frac{z - z_1}{z - z_2} ight) = \frac{\pi}{4}$ represents an arc of a circle passing through $z_1$ and $z_2$.The locus is given by $\frac{(y-6)(x-4) - (y-6)(x-10)}{(x-4)(x-10) + (y-6)^2} = 	an\left(\frac{\pi}{4}ight) = 1$.Simplifying,we get $(y-6)(x-4 - x + 10) = (x-4)(x-10) + (y-6)^2$.$6(y-6) = x^2 - 14x + 40 + y^2 - 12y + 36$.$x^2 - 14x + y^2 - 18y + 76 + 36 = 0 \implies x^2 - 14x + y^2 - 18y + 112 = 0$.Adding $49 + 81$ to both sides to complete the square:$(x^2 - 14x + 49) + (y^2 - 18y + 81) = -112 + 49 + 81 = 18$.$(x - 7)^2 + (y - 9)^2 = 18$.We need to find $|z - 7 - 9i| = |(x-7) + i(y-9)| = \sqrt{(x-7)^2 + (y-9)^2}$.Substituting the value from the equation of the circle,we get $\sqrt{18} = 3\sqrt{2}$.

If ${z_1} = 10 + 6i$,${z_2} = 4 + 6i$ and $z$ is a complex number such that $\text{amp}\left( \frac{z - z_1}{z - z_2} \right) = \frac{\pi}{4}$,then the value of $|z - 7 - 9i|$ is equal to

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