Given two points on the sphere and the circle $k$, which passes through exactly one of the two points. How many circles are there on the sphere that pass through both points and are tangent to $k$?
I. Solution. We will show that exactly one such circle exists.

Let $\mathcal{G}$ be the surface of the sphere, $A$ the point on $k$ where it touches, $B$ the other given point, $\mathcal{S}$ the plane containing $k$, and $e$ the tangent line to the circle $k$ at $A$ in the plane $\mathcal{S}$. Since $\mathcal{G} \cap \mathcal{S}=k$, the points of the line $e$ other than $A$ are outside the sphere, so $e$ is also tangent to the sphere (Figure 1). Therefore, if any plane $\mathcal{T}$ contains $e$ and intersects $\mathcal{G}$ in a circle $\ell$, then the tangent to $\ell$ at $A$ is also $e$, meaning $\ell$ is tangent to $k$ at $A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_437a9535b786f41a453ag-1.jpg?height=551&width=556&top_left_y=364&top_left_x=774)

Figure 1

If we choose the plane $\mathcal{T}$ determined by the point $B$ and the line $e$, then this plane contains two different points of the sphere, so $\mathcal{G} \cap \mathcal{T}$ is a circle, and by the previous arguments, this circle is tangent to $k$ at $A$.

If, however, some circle $m$ is such that the plane $\mathcal{U}$ cutting $m$ from $\mathcal{G}$ passes through both $A$ and $B$ but does not contain $e$, then the line $\mathcal{S} \cap \mathcal{U}=f$ lies in the plane $\mathcal{S}$ and passes through the point $A$ of the circle $k$, but differs from the tangent to $k$ at $A$. Therefore, $f$ intersects $k$ at another point $C$ different from $A$ (Figure 2). This point $C$ is also on the circle $\mathcal{G} \cap \mathcal{U}=m$, so $m$ and $k$ have at least (and easily seen to have exactly) two common points, meaning they do not touch each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_437a9535b786f41a453ag-1.jpg?height=510&width=486&top_left_y=1309&top_left_x=814)

Figure 2

This proves our statement.

II. Solution. Using the notation from the first solution, let $\mathcal{E}$ be the tangent plane to the sphere at the point antipodal to $A$. We solve the problem using a transformation called stereographic projection.

Assign to any point $P$ on the sphere's surface, different from $A$, the point $P^{\prime}=\mathcal{E} \cap A P$ (Figure 3). We will show that this correspondence is a bijection between the sets $\mathcal{G} \backslash\{A\}$ and $\mathcal{E}$. If $P \in(\mathcal{G} \backslash\{A\})$ is any point, then the line $A P$ is not parallel to $\mathcal{E}$, so there exists a unique point $P^{\prime} \in \mathcal{E}$. If $Q^{\prime} \in \mathcal{E}$ is any point, then the line $A Q^{\prime}$ is not in the tangent plane to the sphere at $A$, because that plane is parallel to $\mathcal{E}$, so the line intersects the sphere's surface at two points, hence there exists a unique point $Q$ different from $A$ where $A Q^{\prime}$ intersects $\mathcal{G}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_437a9535b786f41a453ag-1.jpg?height=387&width=593&top_left_y=2273&top_left_x=755)

Figure 3

If $\ell$ is a circle on $\mathcal{G}$ passing through $A$, then the image $\ell^{\prime}$ of $\ell$ under the correspondence is a line, because if $\ell$ is cut from $\mathcal{G}$ by the plane $\mathcal{T}$, then the lines connecting $A$ to the points of $\ell$ lie in $\mathcal{T}$, so $\ell^{\prime}$ is precisely the line $\mathcal{E} \cap \mathcal{T}$ (Figure 4). Conversely, if $m^{\prime}$ is any line in $\mathcal{E}$, then the plane $\mathcal{U}$ determined by $A$ and $m^{\prime}$ intersects the sphere's surface in a circle $m$ passing through $A$, whose image under the correspondence is the line $m^{\prime}$. Therefore, the correspondence is a bijection between the circles on $\mathcal{G}$ passing through $A$ and the lines in $\mathcal{E}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_437a9535b786f41a453ag-2.jpg?height=420&width=593&top_left_y=373&top_left_x=755)

Figure 4

The solution to our problem is now simple. Let the image of $k$ under the correspondence be the line $k^{\prime}$, and the image of $B$ be the point $B^{\prime}$ not on $k^{\prime}$. A circle on the sphere's surface is exactly tangent to $k$ at $A$ if and only if $k \cap n=\{A\}$, which means $k^{\prime} \cap n^{\prime}=\emptyset$, or that the lines $k^{\prime}$ and $n^{\prime}$ are parallel (Figure 5). Exactly one line can be drawn through the point $B^{\prime}$ parallel to $k^{\prime}$, so there is exactly one circle on the sphere that passes through both $A$ and $B$ and is tangent to $k$.

![](https://cdn.mathpix.com/cropped/2024_05_02_437a9535b786f41a453ag-2.jpg?height=198&width=366&top_left_y=1099&top_left_x=869)

Figure 5

Remark. The stereographic projection has many other interesting properties. For example, the image of a circle on the sphere's surface that does not pass through $A$ is a circle in $\mathcal{E}$. The stereographic projection is also frequently used in map-making, where there is a need to represent the sphere's surface on a plane.