Test Index

Complex Numbers Part 2

Section: Mathematics

Question : 1 of 106

Marks: +1, -0

z \in C

|z-(4+8i)| = \sqrt{10} \text{ and } |z-(3+5i)| + |z-(5+11i)| = 4\sqrt{5}

Solution:

|z - (4+8i)| = \sqrt{2}

It is a circle with center (4, 8), radius =

\sqrt{10}

|z - (3+5i)| + |z - (5+11i) = 4\sqrt{5}

It is an ellipse with focii

S(3,5) \& S'(5,11)

SS' = \sqrt{4+36} \Rightarrow \sqrt{40} = 2\sqrt{10}

Clearly

4\sqrt{5} > SS'

Center of ellipse

\left( \frac{3+5}{2}, \frac{5+11}{2} \right) \equiv (4,8)

coincides

with center of circle

Now,

2a = 4\sqrt{5}

a = 2\sqrt{5}

c = \sqrt{10}

b^2 = 20 - 10 = 10

b = \sqrt{10}

= radius of circle

∴ ellipse & circle will intersect at end points of minor axis

So, 2 points of intersection

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