# find minimum value of PQ

Consider the locus of the complex number $$'z'$$ in the argand plane is given by $$Re(z)-2=|z-7+2i|$$.Let P($$z_1$$) and Q($$z_2$$) be two complex numbers satisfying the given locus and also satisfying $$arg\bigg(\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}\bigg)=\dfrac{\pi}{2}$$($$\alpha\in R$$) then the minimum value of PQ(distance)

My try:

let $$z=x+iy$$

from the locus of $$z$$ given,we get $$(x-2)=\bigg|(x-7)+i(y+2)\bigg|$$

$$\implies$$ $$(x-2)^2=(x-7)^2+(y+2)^2$$

$$\implies(y+2)^2=5(2x-9)$$

moving on we have $$2$$nd equation, $$arg\bigg(\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}\bigg)=\dfrac{\pi}{2}$$

which means that the value in the argument lies on positive side of $$y-axis$$

$$\dfrac{z_1-(2+i\alpha)}{z_2-(2+i\alpha)}=ib$$ for $$b>0,b\in R$$ plugging in $$z_1$$=$$x_1+iy_1$$ and $$z_2=x_2+iy_2$$, I got,

$$x_1-2=-b(y_2-\alpha)$$

$$b(x_2-2)=y_1-\alpha$$

how to proceed further$$?$$Any better way to solve equation $$2$$ $$?$$