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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| is less than 1 .

saurabh , 10 Years ago
Grade 12th pass
anser 2 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find the answer to your question below
|z-ai| = a + 4
|z-2| \leq 1

For having atleast one solution for both the equation, the first circle should intersect with 2ndcircle on the x – axis.
|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}
Intersection with x-axis, we have
x^{2}+(0-a)^{2} = (a + 4)^{2}
x^{2} = 8a + 16
8a + 16 = (3)^{2}
8a + 16 = 9
a = \frac{-7}{8}

saurabh

Last Activity: 10 Years ago

first of all |z-2| \leq 1 is not right as given in qus this is less than 1 not equal second thing this qus is from resonannce assignment and ans is (-21/10 ,- 5/6)

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