Since $AM \perp MN$, $AM$ is a diameter of $k_1$. Also, the intersection $Q$ of $AB$ and $MN$ is the midpoint of $MN$ (since it lies on the radical axis of both circles, so the power of it with respect to both circles is the same). Thus, since $MN=2AM$, $\triangle AMQ$ is an isoceles right triangle. Since $AM$ is a diameter, $\angle ABM=90$, so $\angle NMB=45$.

<NMB=45

dgreenb801 wrote: Since $AM \perp MN$, $AM$ is a diameter of $k_1$. Also, the intersection $Q$ of $AB$ and $MN$ is the midpoint of $MN$ (since it lies on the radical axis of both circles, so the power of it with respect to both circles is the same). Thus, since $MN=2AM$, $\triangle AMQ$ is an isoceles right triangle. Since $AM$ is a diameter, $\angle ABM=90$, so $\angle NMB=45$. But AM is the least distance from the point A on MN. Moreover, angle AMN =90. Then how can AQ = AM?? You cannot draw two perpendiculars at two distinct points on one line from a particular point, can you?

Answer is $45$°. Simply notice that the radical axis of the circles bisects the common tangent and you get an isosceles right triangle. Then, apply the alternate segment theorem.

combidude wrote: dgreenb801 wrote: Since $AM \perp MN$, $AM$ is a diameter of $k_1$. Also, the intersection $Q$ of $AB$ and $MN$ is the midpoint of $MN$ (since it lies on the radical axis of both circles, so the power of it with respect to both circles is the same). Thus, since $MN=2AM$, $\triangle AMQ$ is an isoceles right triangle. Since $AM$ is a diameter, $\angle ABM=90$, so $\angle NMB=45$. But AM is the least distance from the point A on MN. Moreover, angle AMN =90. Then how can AQ = AM?? You cannot draw two perpendiculars at two distinct points on one line from a particular point, can you? AMQ is isoceles because AM=MQ

Let $L=\overline{AB}\cap\overline{MN},$ and note $L$ is the midpoint of $\overline{MN}$ by PoP. Then, $AM=ML$ and $\angle AML=90$ so $\angle BMN=\angle BAM=45.$ $\square$

Anyone regretting not knowing the "distance between two points" formula at the time? Let $O_1$ be the center of $k_1$ (the midpoint of $AM$, as $AM \perp MN$) and $O$ be the center of $k_2$ (we have $ON \perp MN$). Choose coordinates so that $M(0,0)$ and $O_1(0,1)$. Then $A(0,2)$ as $O_1$ is the midpoint of $AM$ and $N(4,0)$. The point $O$ has $x$-coordinate $4$, and its $y$-coordinate $r$ is determined from $ON^2 = OA^2$, i.e. $r^2 = 4^2 + (r-2)^2$, giving $r = 5$. For $B(b_1,b_2)$ we have $OB^2 = ON^2$ and $O_1B^2 = O_1M^2$, i.e. $(b_1-4)^2 + (b_2-5)^2 = 5^2$ and $b_1^2 + (b_2-1)^2 = 1$. Equivalently, we have $b_1^2 + b_2^2 - 8b_1 - 10b_2 + 16 = 0 $ and $ b_1^2 + b_2^2 - 2b_2 = 0$. Substituting $b_1^2 + b_2^2$ from the latter into the former gives $b_1 = 2-b_2$. Hence $(2-b_2)^2 + b_2^2 - 2 = 0$, i.e. $2(b_2-1)^2 = 0$, thus $b_1 = b_2 = 1$. Hence $B$ lies on the angle bisector of the first quadrant, therefore $\angle NMB = 45^{\circ}$.

way too easy for jbmo,just note that $AB$ bisects $MN$ and the proof is easy by PoP