I think this is easiest if you use the fact that you can plot any point $(x,y)$ on a circle with center $(a,b)$ using something like the following equations:
\begin{align}
x &= a + r \cos\theta,\\
y &= b + r \sin\theta,
\end{align}
where $\theta$ is the direction angle from the center to the desired point
and $r$ is the radius of the circle.
I say the equations are "something like" this because you might be measuring the direction angle clockwise or counterclockwise, some applications have different "zero angle" directions, and sometimes the axes are different
(for example, the $y$ axis points downward instead of upward).
The equations above are for an angle measured counterclockwise from the positive $x$ axis, with the $x$ axis pointing rightward and the $y$ axis pointing upward.
As long as the angle is measured from one of the axial directions, however, the correct equations are some variation of the above: either as written, or swapping $\sin$ and $\cos$ functions, and/or changing one or the other of the $+$ operators to $-$.
If you can get the correct answers for the angles zero and $90$ degrees then you know you have the correct variation of the equations for your application.
So let's say the "zero" direction angle is from $(a,b)$ to $(X,Y).$
Now consider the triangle with the three vertices $(a,b),$ $(X,Y),$ and $(X',Y').$
It is an isosceles triangle with angle $t$ at each of the vertices
$(X,Y),$ and $(X',Y')$; therefore the angle at $(a,b)$ is
$180 - 2t \text{ degrees}$ (assuming you measured the angle $t$ in degrees).
Not knowing exactly how your application maps coordinates to the plane, I'll assume your $y$ axis points upwards.
But the angle $180 - 2t \text{ degrees}$ would be measured clockwise in the figure you drew.
You have stated that $a = 0$ and $b = 0$ in this figure.
Also, most computer libraries with sine and cosine functions expect the angle to be given in radians rather than degrees.
Under these assumptions, the particular equations to find the point you want would be
\begin{align}
X' &= r \cos\theta,\\
Y' &= -r \sin\theta
\end{align}
where $\theta$ is $180 - 2t$ converted from degrees to radians.
