Geometry (Locus and constructions) 
*

*I want to find the equation for the locus that is at the same distance from the point $(2,3)$ to the line $x=1$. 
Im not sure if I am right or wrong? Is the locus just the two point at a distance=1 above and below the point $(2,3)$? 

*Two point A and B forms a segment $AB= 4$cm. Any point $P$ such at angle( APB) = $30$ lies on a circle through A and B. Construct the circle that P has to lie on. 


I appreciate any answers, I am stuck.
 A: The general method to solve such kind of problems is to take a point $P=(x,y)$ and write the given conditions. 
For 1) we have the point $F=(2,3)$ and the line $d$ of equation $x=1$  and we want that the distance $\overline {Pd}$ is equal to the distance $\overline{PF}$. You can easily that $\overline {Pd}=|x-1|$ and, using the formula for the distance : $\overline{PF}=\sqrt{(x-2)^2+(y-3)^2}$, so the coordinates of the point $P$ must satisfie the equation
$$
|x-1|=\sqrt{(x-2)^2+(y-3)^2}
$$  
squaring and reordering you can see that this is the equation of a parabola.
For 2), without loss of generality, you can take $A=(0,0)$ and $B=(4,0)$ and  $P=(x,y)$. Now use the fact that $PA$ has slope $\frac{y}{x}= \tan \alpha$ and $PB$ has slope $\frac{y}{x-4}=\tan \beta$, and $\beta-\alpha=30°$.
So we have:
$$
\tan 30°=\frac{1}{\sqrt{3}}=\tan( \beta-\alpha)=\frac{\tan \beta -\tan \alpha}{1+\tan \beta \tan \alpha}
$$
substituting $\tan \alpha$ and $\tan \beta$ and reducing the equation you find the equation of a circle and you can prove that the points $A$ and $B$ are points of this circle.

1) squaring the equation we have:
$$
x^2+1-2x=x^2+4-4x+y^2+9-6y \quad \iff \quad 2x=y^2-6y+12 \quad \iff \quad x=\frac{1}{2}y^2-3y+6
$$
that is the equation of a parabola with the symmetry axis parallel to the $x$ axis.  Note that this exercise is exactly the definition of a parabola with directrix the line $d$ and focus the point $F$.
