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Find the real values of the parameter ‘a’ for which atleast one complex number z =x+iy satisfies both the equality |z- ai|=a+4 and the inequality | z-2|

saurabh , 10 Years ago
Grade 12th pass
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question below
|z-ai| = a + 4
|z-2| < 1
For these two to have atleast one solution, then both circles must intersect
|x+iy-ai| = a+4
|x+i(y-a)| = a + 4
|x+i(y-a)|^{2} = (a + 4)^{2}
x^{2}+(y-a)^{2} = (a + 4)^{2}
1st circle intersect the x – axis at 1 & 3.
2nd circle intersect x – axis
x^{2}+(0-a)^{2} = (a + 4)^{2}
x^{2} + a^{2} = a^{2} + 16 + 8a
x^{2} = 16 + 8a
16 + 8a = 1
a = \frac{-15}{8}
16 + 8a = 9
a = \frac{-7}{8}
\Rightarrow \frac{-15}{8}< a< \frac{-7}{8}

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