Show that a circle on the sphere through the north pole…
Show that a circle on the sphere through the north pole $(0,0,1)$ projects (stereographically) to a line in the $z$-plane, and that every line or circle in the $z$-plane is the image of some circle on the sphere.
Explanation of Method
Every circle on the sphere is the intersection of the sphere with a plane $A\xi+B\eta+C\zeta=D$. Substitute the stereographic projection formulas $\xi=\frac{x(1-\zeta)}2$, $\eta=\frac{y(1-\zeta)}2$ into this plane equation; the $(1-\zeta)$ factor cancels (since $\zeta\neq1$ off the north pole), collapsing the equation to one purely in $x,y$ — linear if the plane passes through the north pole, quadratic (a circle) otherwise.
Part 1 · Circles through the north pole become lines
Step 1 · A circle on the sphere is a plane section
Every circle on the unit sphere is $\{(\xi,\eta,\zeta):\xi^2+\eta^2+\zeta^2=1\}\cap\{A\xi+B\eta+C\zeta=D\}$ for some real $A,B,C,D$ (not all of $A,B,C$ zero).
Step 2 · Impose passage through $(0,0,1)$
Substituting $(0,0,1)$: $\ C=D$. The plane becomes $A\xi+B\eta+C(\zeta-1)=0$.
Step 3 · Substitute the projection formulas
From $z=\dfrac{2(\xi+i\eta)}{1-\zeta}$: $\ \xi=\dfrac{\text{Re}(z)(1-\zeta)}2,\ \eta=\dfrac{\text{Im}(z)(1-\zeta)}2$. Substituting:
$$ A\cdot\frac{\text{Re}(z)(1-\zeta)}2+B\cdot\frac{\text{Im}(z)(1-\zeta)}2+C(\zeta-1)=0 $$
Factor out $(1-\zeta)$ (using $\zeta-1=-(1-\zeta)$):
$$ (1-\zeta)\left[\frac{A\,\text{Re}(z)+B\,\text{Im}(z)}2-C\right]=0 $$
Since $\zeta\neq1$ (we are away from the north pole):
$$ \boxed{A\,\text{Re}(z)+B\,\text{Im}(z)=2C} $$
a line in the $z$-plane.
Part 2 · Every line or circle in the plane comes from a sphere circle
Step 4 · General plane-curve equation
Every line or circle in $\mathbb C$ can be written as $\ \alpha|z|^2+\beta\,\text{Re}(z)+\gamma\,\text{Im}(z)+\delta=0\ $ ($\alpha=0$: a line; $\alpha\neq0$: a circle).
Step 5 · Substitute the sphere coordinates
Using $|z|^2=\dfrac{4(1+\zeta)}{1-\zeta}$, $\ \text{Re}(z)=\dfrac{2\xi}{1-\zeta}$, $\ \text{Im}(z)=\dfrac{2\eta}{1-\zeta}$, then multiplying through by $(1-\zeta)$:
$$ 4\alpha(1+\zeta)+2\beta\xi+2\gamma\eta+\delta(1-\zeta)=0 $$
$$ \boxed{2\beta\xi+2\gamma\eta+(4\alpha-\delta)\zeta=-(4\alpha+\delta)} $$
which is exactly a plane equation $A\xi+B\eta+C\zeta=D$. Its intersection with the sphere is a circle.
Verification
Take the line $\text{Re}(z)=1$ ($\alpha=0,\beta=1,\gamma=0,\delta=-1$). Step 5 gives the sphere plane $2\xi+\zeta=1$. Checking two points on the line: $z=1$ has sphere image $\left(\frac45,0,-\frac35\right)$ (from Problem 42's formulas), and indeed $2\cdot\frac45+\left(-\frac35\right)=\frac85-\frac35=1$ ✓. Likewise $z=1+i$ has sphere image $\left(\frac23,\frac23,-\frac13\right)$, and $2\cdot\frac23+\left(-\frac13\right)=\frac43-\frac13=1$ ✓. Both confirm the plane equation from Step 5.
Result
Stereographic projection is circle-preserving in the strongest sense: treating lines as "circles through infinity," every circle on the sphere corresponds to a circle or line in the plane, and vice versa.