homework and exercises - How to get image current in magnetostatics? (magnetic slab)

I want to know the image current and its location which satisfies the boundary condition at the interface.

This problem was originated from the problem 6-33 in Fields and wave electromagnetics, D. Cheng, 2nd Ed.

To solve this problem, the following is what i did.

With equations in magnetostatics and its boundary condition,

$$\mathbf{B}=\mu\mathbf{H}, \qquad B_1=B_2 = \mu_1H_{1n}=\mu_2H_{2n}, \qquad H_{t1}=H_{t2}$$ In this case, the normal boundary condition becomes $$H_{1n}=\mu_rH_{2n}$$

From the magnetic flux density which I got from the figure, $\mathbf{B}=\frac{\mu_0I}{2\pi(y^2+d^2)}(y\mathbf{\hat{x}}+x\mathbf{\hat{y}})$, normal and tangential magnetic flux density($\mathbf{B}$) and magnetic field intensity($\mathbf{H}$) at interface adjacent to magnetic medium 1 are

$$(\mathbf{\hat{x}})\quad B_{1n}=\frac{\mu_0Iy}{2\pi(y^2+d^2)}, \qquad H_{1n}=\frac{Iy}{2\pi(y^2+d^2)}$$ $$(\mathbf{\hat{y}})\quad B_{1t}=\frac{\mu_0Ix}{2\pi(y^2+d^2)}, \qquad H_{1t}=\frac{Ix}{2\pi(y^2+d^2)}$$ with this, we can get $\mathbf{B}$ and $\mathbf{H}$ in magnetic medium 2. $$\to H_{2n}=\frac{Iy}{\mu_r2\pi(y^2+d^2)} \to B_{2n}=\frac{\mu_0Iy}{2\pi(y^2+d^2)}$$ $$\to H_{2t}=\frac{Ix}{2\pi(y^2+d^2)} \to B_{2t}=\frac{\mu_0\mu_rIx}{2\pi(y^2+d^2)}$$ Finally, $$\mathbf{B}_2=\frac{\mu_0I}{2\pi(y^2+d^2)}(y\mathbf{\hat{x}}+\mu_rx\mathbf{\hat{y}})$$ Is this whole procedure right? From this, I cannot infer image current $I_i=\left(\frac{\mu_r-1}{\mu_r+1}\right)I$ as stated in the problem. What makes to satisfy boundary condition? How do I get the image current?