Expert Help with High School Geometry

I am trying to figure out what the best method is to go about finding this locus.
$\arg <!--\; \; -->{\displaystyle \frac{z-<!--\; -\; -->a}{z-<!--\; -\; -->b}}=\mathrm{\theta <!--\; \theta \; -->}$
I am aware that it must be part of an arc of a circle that passes through the points $a$ and $b$. The argument means that the vector $(z-<!--\; -\; -->a)$ leads the vector $(z-<!--\; -\; -->b)$ by $\mathrm{\theta <!--\; \theta \; -->}$ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to
$\arg <!--\; \; -->{\displaystyle \frac{z-<!--\; -\; -->a}{z-<!--\; -\; -->b}}=-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}$
Would this just be the other part of the same circle as previously or a different circle?

Perhaps you could help by showing my how it would work with the question
$\arg <!--\; \; -->\frac{z-<!--\; -\; -->2j}{z+3}=\pi /3$

I need help with the following optimization problem

$max\mathrm{\alpha <!--\; \alpha \; -->}\ln <!--\; \; -->(x(1-<!--\; -\; -->y^2))+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})\ln <!--\; \; -->(z)$
where the maximization is with respect to $x,y,z$, subject to
$\begin{array}{rl}\mathrm{\alpha <!--\; \alpha \; -->}x+(1-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})z& =C_1\\ \mathrm{\alpha <!--\; \alpha \; -->}y\sqrt{x(x+\mathrm{\gamma <!--\; \gamma \; -->})}-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}x& =C_2\end{array}$
where $0\le <!--\; \le \; -->\mathrm{\alpha <!--\; \alpha \; -->}\le <!--\; \le \; -->1$, $\mathrm{\gamma <!--\; \gamma \; -->}>0$, and $x,z\ge <!--\; \ge \; -->0$, and $|y|\le <!--\; \le \; -->1$.

Generally, one can substitute the constraints in the objective function and maximize with respect to one parameter. The problem is that in this way things become algebraically complicated, and I believe that there is a simple solution.

Prove that the product of two lines equations is hyperbola

Convert hyperbola in rectangular form to polar form
${\displaystyle 3y^2-16y-x^2+16=0.}$

Let L :${\displaystyle :P_2\to P_3}$ be a linear transformation for which we
know that ${\displaystyle L:\left(1\right)=1,L\left(t\right)=t^2,L\left(t^2\right)=t^3=t.}$
(a) Find ${\displaystyle L(2t^2-5t=3).}$
(b) Find ${\displaystyle L(a{t}^2-bt+c).}$

Can it be shown that gives a conic section?
${\displaystyle r^2=\frac{1}{1-\epsilon \cos \left(2\theta \right)}}$

Assume T: ${\displaystyle R\wedge m}$ to ${\displaystyle R\wedge n}$ is a matrix transformation with matrix A.
Prove that if the columns of A are linearly independent then T is one to one. (i.e injective) (Hint: Remember the matrix transformations satisfy the linearity properties.)
Linearity Properties:
If A is a matrix, v and w are vectors and c is a scalar then,
A0=0
A(cv)=cAv