### Instruction: The parabola $y = x^2 + bx + c$ has the following properties:

The point on the parabola closest to $(12,3)$ is the $y$-intercept of the parabola.
The parabola passes through $(-5,0).$

Enter the ordered pair $(b,c).$.### Response: This problem involves finding the equation of a parabola that satisfies some given conditions.
To start, I notice that the $y$-intercept of the parabola is the point where $x = 0$, so $y = c$ at that point.
The problem says that this point is the closest to $(12,3)$, so I want to minimize the distance between these two points.
The distance between $(0,c)$ and $(12,3)$ is given by the formula $d = \sqrt{(12-0)^2 + (3-c)^2}$.
To minimize $d$, I want to minimize the expression under the square root, which is $(12-0)^2 + (3-c)^2 = 144 + (3-c)^2$.
This is a quadratic function of $c$, and it has a minimum value when $c$ is the average of the roots.
The roots of this function are the values of $c$ that make $(3-c)^2 = 0$, which are $c = 3$ and $c = 3$.
So the average of the roots is $\frac{3+3}{2} = 3$, and this is the value of $c$ that minimizes $d$.
Therefore, the $y$-intercept of the parabola is $(0,3)$.
Now I have one point on the parabola, and I need to find $b$.
I also know that the parabola passes through $(-5,0)$, so I can plug in those values into the equation and get $0 = (-5)^2 + b(-5) + 3$.
Simplifying, I get $25 - 5b + 3 = 0$, or $5b = 28$.
Dividing by $5$, I get $b = \frac{28}{5}$.
So the equation of the parabola is $y = x^2 + \frac{28}{5}x + 3$.
The ordered pair $(b,c)$ is $(\frac{28}{5},3)$.