If I understood your question correctly, the differences $\Delta x$, $\Delta y$, $\Delta z$ for correction are likely equivalent to the round up of your coordinates namely:
$$\Delta x=+0.04$$
$$\Delta y=+0.01$$
$$\Delta z=+0.1$$
If you must have the correction in spherical coordinates, further you may take the general transformation rules for line element (approximating small differences $\Delta$) you have the Jacobi matrix:
$$\frac{\Delta(x,y,z)}{\Delta(r,\theta,\varphi)}\simeq J=\begin{pmatrix}
\sin\theta\cos\varphi &r\cos\theta\cos\varphi&-r\sin\theta\sin\varphi\\
\sin\theta \sin\varphi&r\cos\theta\sin\varphi&r\sin\theta\cos\varphi\\
\cos\theta&-r\sin\theta&0
\end{pmatrix}$$
and:
$$(\Delta x,\Delta y,\Delta z)^T=J\cdot(\Delta r,\Delta \theta,\Delta \varphi)^T$$
See also here>>>; The system of equation must be solved. You may also need to take into consideration:
$$r=\sqrt{x^2+y^2+z^2}$$
You can compute the homography matrix H with your eight points with a matrix system such that the four correspondance points $(p_1, p_1'), (p_2, p_2'), (p_3, p_3'), (p_4, p_4')$ are written as $2\times9$ matrices such as:
$p_i = \begin{bmatrix}
-x_i \quad -y_i \quad -1 \quad 0 \quad 0 \quad 0 \quad x_ix_i' \quad y_ix_i' \quad x_i' \\
0 \quad 0 \quad 0 \quad -x_i \quad -y_i \quad -1 \quad x_iy_i' \quad y_iy_i' \quad y_i'
\end{bmatrix}$
It is then possible to stack them into a matrix $P$ to compute:
$PH = 0$
Such as:
$PH = \begin{bmatrix}
-x_1 \quad -y_1 \quad -1 \quad 0 \quad 0 \quad 0 \quad x_1x_1' \quad y_1x_1' \quad x_1' \\
0 \quad 0 \quad 0 \quad -x_1 \quad -y_1 \quad -1 \quad x_1y_1' \quad y_1y_1' \quad y_1' \\
-x_2 \quad -y_2 \quad -1 \quad 0 \quad 0 \quad 0 \quad x_2x_2' \quad y_2x_2' \quad x_2' \\
0 \quad 0 \quad 0 \quad -x_2 \quad -y_2 \quad -1 \quad x_2y_2' \quad y_2y_2' \quad y_2' \\
-x_3 \quad -y_3 \quad -1 \quad 0 \quad 0 \quad 0 \quad x_3x_3' \quad y_3x_3' \quad x_3' \\
0 \quad 0 \quad 0 \quad -x_3 \quad -y_3 \quad -1 \quad x_3y_3' \quad y_3y_3' \quad y_3' \\
-x_4 \quad -y_4 \quad -1 \quad 0 \quad 0 \quad 0 \quad x_4x_4' \quad y_4x_4' \quad x_4' \\
0 \quad 0 \quad 0 \quad -x_4 \quad -y_4 \quad -1 \quad x_4y_4' \quad y_4y_4' \quad y_4' \\
\end{bmatrix} \begin{bmatrix}h1 \\ h2 \\ h3 \\ h4 \\ h5 \\ h6 \\ h7 \\ h8 \\h9 \end{bmatrix} = 0$
While adding an extra constraint $|H|=1$ to avoid the obvious solution of $H$ being all zeros. It is easy to use SVD $P = USV^\top$ and select the last singular vector of $V$ as the solution to $H$.
Note that this gives you a DLT (direct linear transform) homography that minimizes algebraic error. This error is not geometrically meaningful and so the computed homography may not be as good as you expect. One typically applies nonlinear least squares with a better cost function (e.g. symmetric transfer error) to improve the homography.
Once you have your homography matrix $H$, you can compute the projected coordinates of any point $p(x, y)$ such as:
$\begin{bmatrix}
x' / \lambda \\
y' / \lambda \\
\lambda
\end{bmatrix} =
\begin{bmatrix}
h_{11} & h_{12} & h_{13} \\
h_{21} & h_{22} & h_{23} \\
h_{31} & h_{32} & h_{33}
\end{bmatrix}.
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}$
Best Answer
I found the answer. Mathematically there is no way to do what I asked for. but by adding some other information it is reachable. in this context: (I assumed a projective transformation as I was working with camera modeling and 3D image processing) either of following can give us enough information to compute other H matrix:
C :camera position
Vz : vanishing point in Z direction
etc... (there are other measurable parameters that also work)