Let $BC=2,BM=1,I$ is the incenter of the circle. Now, $BK=\sqrt{5}$ $MK$ is diameter of the circle. So, $ML\perp BL$ or, $MK\perp BK$. Area of $\triangle BMK=\frac{1}{2}\times BM\times MK=\frac{1}{2}\times ML\times BK$ $\implies 1\times 2=ML\times \sqrt{5}$ $\implies ML=\frac{2}{\sqrt{5}}$ So, $KL=\sqrt{(MK^2)-(ML^2)}$ $\implies KL=\frac{4}{\sqrt{5}}$ So, $KX=XL=\frac{2}{\sqrt{5}}=ML$ So, $\triangle MKL$ is an isosceles right angled triangle. $\implies \angle MXL=45^{\circ}$ $\implies \angle MXK=180^{\circ}-45^{\circ}=\boxed{135^{\circ}}$

Clearly $X$ is the foot from $O$ to $BK$. Then OMBX is cyclic, i.e. <OXM = <OBM = 45, so <MXK = 45+90=135

Clearly $X$ is the foot from $O$ to $BK$. Then $OMBX$ is cyclic, i.e. $\angle OXM = \angle OBM = 45^{\circ}, so \angle MXK = 45^{\circ}+90^{\circ}=135^{\circ}$