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My working:

$$ \frac{x + iy - 2}{x + iy + 5} $$ $$ \frac{(x - 2 + iy)(x+5-iy)}{(x + 5 + iy)(x+5-iy)} $$ $$ \frac{x^2+5x-ixy-2x-10+2iy+ixy+5iy+y^2}{x^2+5x-ixy+5x+25-5iy+ixy+5iy+y^2} $$ $$ \frac{x^2+3x-10+y^2+7iy}{x^2+10x+25+y^2}$$ $$ \frac {\Im(z)}{\Re(z)} = \tan \frac{\pi}{4} = 1$$ $$ x^2 + 3x + y^2 - 10 = 7y $$ $$ x^2 + 3x + y^2 - 7y = 10 $$ $$ \left(x+ \frac 32\right)^2 + \left(y- \frac 72\right)^2 = 10 + \frac 94 + \frac {49}{4} $$ $$ \left(x- -\frac 32\right)^2 + \left(y- \frac 72\right)^2 = \left(\frac {7 \sqrt{2}}{2}\right)^2 $$

Is this correct? Is there a better method than what I did here?

UmbQbify
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Wooneh
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2 Answers2

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Let me try this way -

The numerator is the line joining $Z(x,y)$ to $(2,0)$ while the denominator joins $Z(x,y)$ to $(-5,0)$ .

Geometrically, These two points are at a $45^\circ$ angle to each other.

Both the points should be on the locus. (Imagine a very small line at each point making $45^\circ$ angle with the other line)

Now, two points on the locus suspend an angle of $45^\circ$ to any other point on the locus implies - the locus is a circle!! and the center is at $90^\circ$ to these two line.

The center also lies on the perpendicular bisector of these two points, and so, the center is at $(-1.5,3.5)$. The two points are on the circle. So, radius is done too.

Michael Hardy
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Win Vineeth
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Hint for another method:

Translate this problem into pure geometry: denote $A$ the point with affix $-5$, $B$ the point with affix $2$, $M$ the point with affix $z$. The condition translates into $\angle AMB=\frac\pi4$.

Now use the inscribed angle theorem: the locus of $M$ is a circular arc. The centre $O$ of this arc is such that $\angle AOB=\frac\pi2$. Hence $O$ is the intersection point of the upper semi-circle with diameter $[AB]$ and the perpendicular bisector of $[AB]$.

Bernard
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  • Well, I was a 4 seconds late, but I solved it completely, so I am not removing my answer. – Win Vineeth Mar 12 '16 at 01:14
  • The locus is only an arc of the circle, if I'm not mistaken. – Bernard Mar 12 '16 at 01:17
  • @WinVineeth : The fact that your answer is not the first one or not the only one is not a reason to delete it. $\qquad$ – Michael Hardy Mar 12 '16 at 01:17
  • @WinVineeth: I agree with Michael Hardy. – Bernard Mar 12 '16 at 01:18
  • @MichaelHardy Oh well, I thought we can't put the same answer, once someone already did that. Many a time, people deleted their posts because the same answer has already been given. This is to be discussed in meta, not here anyway. – Win Vineeth Mar 12 '16 at 01:48