If at least one value of the complex number z | vedclass

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(A) The condition $|z + \sqrt{2}| = a^2 - 3a + 2$ represents a circle with center $A(-\sqrt{2}, 0)$ and radius $R_1 = \sqrt{a^2 - 3a + 2}$.For $R_1$ t

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$a > 2$

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(A) The condition $|z + \sqrt{2}| = a^2 - 3a + 2$ represents a circle with center $A(-\sqrt{2}, 0)$ and radius $R_1 = \sqrt{a^2 - 3a + 2}$.For $R_1$ to be defined,$a^2 - 3a + 2 \ge 0$,which implies $(a-1)(a-2) \ge 0$,so $a \le 1$ or $a \ge 2$.The inequality $|z + i\sqrt{2}| The distance between centers $A$ and $B$ is $d = \sqrt{(-\sqrt{2} - 0)^2 + (0 - (-\sqrt{2}))^2} = \sqrt{2 + 2} = 2$.For the two regions to have at least one common point,the distance $d$ must be less than the sum of the radii: $d $2 $2 - a If $2 - a  2$),the inequality holds since the $RHS$ is positive.If $2 - a \ge 0$ (i.e.,$a \le 2$),squaring both sides gives $4 - 4a + a^2  2$. This contradicts $a \le 2$.Thus,the condition is satisfied for $a > 2$.

If at least one value of the complex number $z = x + iy$ satisfies the condition $|z + \sqrt{2}| = a^2 - 3a + 2$ and the inequality $|z + i\sqrt{2}| < a^2$,then

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