If the point P denotes the complex number z = x + iy in the Argand plane and \; z-(2-i)/z+(1+2i) is purely imaginary number, then the locus of P is [AP EAPCET 24 May 2025 S1]

Correct Answer is : a circle not containing the point (−1,−2)(-1,-2)(−1,−2) and having its centre on the line x+y+1=0x+y+1=0x+y+1=0

a hyperbola not containing the point (-1,-2)

an ellipse not containing the point (-1,-2)

a parabola not containing the point (-1,-2)

a circle not containing the point (-1,-2) and having its centre on the line x+y+1=0

Correct Answer is : a circle not containing the point (−1,−2)(-1,-2)(−1,−2) and having its centre on the line x+y+1=0x+y+1=0x+y+1=0

Test Index

Complex Numbers

Section: Mathematics

Question : 1 of 50

Marks: +1, -0

z = x + iy

\; \frac{z-(2-i)}{z+(1+2i)}

P

a hyperbola not containing the point $(-1,-2)$
an ellipse not containing the point $(-1,-2)$
a parabola not containing the point $(-1,-2)$
a circle not containing the point $(-1,-2)$ and having its centre on the line $x+y+1=0$

Solution:

We note that the quotient

\; \frac{z-(2-i)}{z+(1+2i)}

is not defined if

z = -(1+2i) \Leftrightarrow x = -1

and

y = -2

Since

x = x + iy

, then

\;\; \frac{z-(2-i)}{z+(1+2i)} = \; \frac{x+iy-(2-i)}{x+iy+(1+2i)}

\; = \; \frac{(x-2)+i(y+1)}{(x+1)+i(y+2)}

\; = \; \frac{(x-2)+i(y+1)}{(x+1)+i(y+2)} \times \; \frac{(x+1)-i(y+2)}{(x+1)-i(y+2)}

\; = \; \frac{(x-2)(x+1)-i(x-2)(y+2)}{(y+1)(y+2)+i(y+1)(x+1)}

\; = \; \frac{(x-2)(x+1)+(y+1)(y+2)}{(x+1)^2+(y+2)^2} + \; \frac{i[(x+1)(y+1)-(x-2)(y+2)]}{(x+1)^2+(y+2)^2}

Since,

\; \frac{z-(2-i)}{z+(1+2i)}

is purely imaginary,

So real part

=0

\; \therefore \; \frac{(x-2)(x+1)+(y+1)(y+2)}{(x+1)^2+(y+2)} = 0

\; \Leftrightarrow x^2+x-2x-2+y^2+2y+y+2=0

\; x^2+y^2-x+3y=0

And

x \neq -1

and

y \neq -2

This is the locus of point

P

. This is a circle.

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