Answer
We show that $f$ is minimized when $c$ is the average of the $x$-coordinates ${a_i}$ and $d$ is the average of the $y$-coordinates ${b_i}$.
Work Step by Step
Let $P = \left( {x,y} \right)$ and $n$ fixed points be $\left( {{a_1},{b_1}} \right),...,\left( {{a_n},{b_n}} \right)$.
Thus, the sum of the squares of the distances from $P$ to these points is given by
$f\left( {x,y} \right) = {\left( {x - {a_1}} \right)^2} + ... + {\left( {x - {a_n}} \right)^2} + {\left( {y - {b_1}} \right)^2} + ... + {\left( {y - {b_n}} \right)^2}$
The partial derivatives are
${f_x} = 2\left( {x - {a_1}} \right) + ... + 2\left( {x - {a_n}} \right) = 2nx - 2\left( {{a_1} + ... + {a_n}} \right)$
${f_y} = 2\left( {y - {b_1}} \right) + ... + 2\left( {y - {b_n}} \right) = 2ny - 2\left( {{b_1} + ... + {b_n}} \right)$
Now, we find the critical points of $f$ by solving ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 2nx - 2\left( {{a_1} + ... + {a_n}} \right) = 0$
${f_y} = 2ny - 2\left( {{b_1} + ... + {b_n}} \right) = 0$
So, there is one critical point $\left( {x,y} \right) = \left( {\frac{{{a_1} + ... + {a_n}}}{n},\frac{{{b_1} + ... + {b_n}}}{n}} \right)$.
We notice that $\frac{{{a_1} + ... + {a_n}}}{n}$ is the average of the $x$-coordinates ${a_i}$ and $\frac{{{b_1} + ... + {b_n}}}{n}$ is the average of the $y$-coordinates ${b_i}$.
Write $c = \frac{{{a_1} + ... + {a_n}}}{n}$ and $d = \frac{{{b_1} + ... + {b_n}}}{n}$, so $c$ is the average of the $x$-coordinates ${a_i}$ and $d$ is the average of the $y$-coordinates ${b_i}$.
Next, we show that $f$ is minimized at the critical point $P = \left( {c,d} \right)$.
The second partial derivatives of $f$ are
${f_{xx}} = 2n$, ${\ \ \ }$ ${f_{yy}} = 2n$, ${\ \ \ }$ ${f_{xy}} = 0$
Since $n > 0$, so ${f_{xx}} = 2n > 0$; and the discriminant $D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 4{n^2} > 0$. By Theorem 2, $f\left( {c,d} \right)$ is a local minimum, Hence, $f$ is minimized when $c$ is the average of the $x$-coordinates ${a_i}$ and $d$ is the average of the $y$-coordinates ${b_i}$.
