For real numbers $a, b, c, d$, the following equations hold:

$$
4 a^{2}+b^{2}-8 b+12=0
$$

and

$$
c^{2}-8 c+4 d^{2}+12=0
$$

What is the minimum value of the expression $(a-c)^{2}+(b-d)^{2}$?
The two equations can be brought to the following form:

$$
\begin{gathered}
a^{2}+\frac{(b-4)^{2}}{4}=1, \quad \text { and } \\
\frac{(c-4)^{2}}{4}+d^{2}=1
\end{gathered}
$$

Equation (1) can be considered as the equation of an ellipse centered at $(0 ; 4)$ with semi-major and semi-minor axes of 1 and 2 units, respectively, while (2) is the equation of an ellipse centered at $(4;0)$ with semi-major and semi-minor axes of 2 and 1 units, respectively. Therefore, the question of the problem is to determine the minimum square of the distance between the points $A(a ; b)$ and $B(c ; d)$ of the ellipses given by

$$
\begin{gathered}
x^{2}+\frac{(y-4)^{2}}{4}=1 \quad \text { and } \\
\frac{(x-4)^{2}}{4}+y^{2}=1
\end{gathered}
$$

where $A$ is a point on (3) and $B$ is a point on (4).

![](https://cdn.mathpix.com/cropped/2024_05_02_1b71ee6d15a573339c7ag-1.jpg?height=723&width=643&top_left_y=831&top_left_x=730)

The two ellipses are symmetric with respect to the line $y=x$, so the minimum distance between their points will be the distance between the two closer tangents that are parallel to the line of symmetry. These tangents are the lines $e_{1}$ and $e_{2}$ in the figure, whose equations are of the form $y=x-t$. If the line $y=x-t$ is a tangent, then the system of equations formed from this equation and, for example, (4) will have exactly one solution, meaning that the equation

$$
\frac{(x-4)^{2}}{4}+(x-t)^{2}=1
$$

will have a discriminant of zero. This equation is:

$$
5 x^{2}-8(t+1) x+\left(4 t^{2}+12\right)=0
$$

and its discriminant is zero:

$$
\begin{gathered}
64(t+1)^{2}-4 \cdot 5\left(4 t^{2}+12\right)=0 \\
t^{2}-8 t+11=0 \\
t=4 \pm \sqrt{5}
\end{gathered}
$$

The value of $t$ corresponding to the tangent closer to the line of symmetry is clearly $t=4-\sqrt{5}$.

Since the line of symmetry and the tangent $e_{2}$ form a $45^{\circ}$ angle with the x-axis, the distance between these two lines is

$$
\begin{gathered}
\frac{4-\sqrt{5}}{\sqrt{2}},(\approx 1.247), \quad \text { so the minimum of }(a-c)^{2}+(b-d)^{2} \text { is: } \\
\left(\frac{2(4-\sqrt{5})}{\sqrt{2}}\right)^{2}=42-16 \sqrt{5}=2.495^{2}
\end{gathered}
$$