Find the Cartesian equation of the locus described by $\arg \left(\frac{z-2}{z+5} \right)= \frac{\pi}{4}$ 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?
 A: 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.
A: 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 $[AB]$.
