The largest value of r for which the region r | vedclass

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(A) The set $\{ \omega \in \mathbb{C} : |\omega - (4 + i)| \le r \}$ represents a disk with center $C(4, 1)$ and radius $r$.The set $\{ z \in \mathbb{

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$\frac{5}{2}\sqrt{2}$

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(A) The set $\{ \omega \in \mathbb{C} : |\omega - (4 + i)| \le r \}$ represents a disk with center $C(4, 1)$ and radius $r$.The set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ represents the region on one side of the perpendicular bisector of the line segment joining $1$ (or $(1, 0)$) and $-i$ (or $(0, -1)$).The perpendicular bisector is the line $x + y = 0$. The region $|z - 1| \le |z + i|$ corresponds to $x + y \ge 0$.For the disk to be contained in the region $x + y \ge 0$,the distance from the center $C(4, 1)$ to the line $x + y = 0$ must be at least $r$.The distance $d$ from $(4, 1)$ to $x + y = 0$ is given by $d = \frac{|4 + 1|}{\sqrt{1^2 + 1^2}} = \frac{5}{\sqrt{2}} = \frac{5}{2}\sqrt{2}$.Thus,the largest value of $r$ is $\frac{5}{2}\sqrt{2}$.

The largest value of $r$ for which the region represented by the set $\{ \omega \in \mathbb{C} : |\omega - 4 - i| \le r \}$ is contained in the region represented by the set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ is equal to

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