The parabola $y = x^2 + bx + c$ has the following properties:
How can I find $(b, c)$?
Here is my attempt:
The point $(0, c)$ is the intersection with the $y$ axis of the parabola. The distance from $(12, 3)$ is $\sqrt{144 + (c-3)^2}$, and we have the equation $-5b +25+c =0$.
But if we don't know vertex of the parabola how do we find $b$ & $c$?
On
The straightforward approach to finding the parabola would be to show the circumstances for which the perpendicular distance from $ \ (12 \ , \ 3 ) \ $ to points on the parabola is smallest for the parabola's $ \ y-$intercept (this is related to Mark Bennet's comment). We can make short work of this with the tools of differential calculus, but will have to go to a bit of trouble earlier in our course of study.
[Note that this is not the same as the question of which parabola has its $ \ y-$intercept closest to $ \ (12 \ , \ 3) \ : \ $ that we could answer immediately as the one with $ \ c \ = \ 3 \ \ $ by using the distance expression you've written.]
The "upward-opening" parabola passing through $ \ (-5 \ , \ 0) \ $ with $ \ y-$intercept $ \ (0 \ , \ c ) \ $ is $ \ y \ = \ x^2 + \left(5 + \frac{c}{5} \right)·x + c \ \ , $ as your work already shows. We will approach the issue of obtaining a "perpendicular distance" by constructing the circle centered on $ \ (12 \ , \ 3) \ $ which has a $ \ y-$intercept at $ \ (0 \ , \ c ) \ \ , \ \ $ the equation for which is $ \ (x - 12)^2 + (y - 3)^2 \ = \ 144 + (c - 3)^2 \ \ . $ The circle and parabola can intersect more than once; this means that there are points on the parabola closer to $ \ (12 \ , \ 3) \ $ than its $ \ y-$intercept is. We wish to find the value of $ \ c \ $ (if it exists) for which the two curves meet only once. It is there that the parabola is tangent to the circle, which makes the radius of the circle perpendicular to that point, and so $ \ (0 \ , \ c ) \ $ is the closest point on the parabola to $ \ (12 \ , \ 3) \ \ . $
We can illustrate this for the case of $ \ c \ = \ 1 \ \ . $ It is clear that the circle has intercepts $ \ y \ = \ 3 \pm 2 \ = \ 1 \ , \ 5 \ \ , $ so it must "pass to the left" of the parabola and intersect it at a second point. To find the coordinates of this second point, we would need to solve the (rather daunting) intersection equation obtained from $ \ (x - 12)^2 + (y - 3)^2 \ \ = \ \ 144 + (1 - 3)^2 \ \ , $ where $ \ y \ = \ x^2 + \left(5 + \frac{1}{5} \right)·x + 1 \ \ , $ or $$ x^2 \ - \ 24 x \ + \ y^2 \ - \ 6y \ \ = \ \ -5 \ \ \ \rightarrow \ \ \ 25·x^4 \ + \ 260·x^3 \ + \ 601·x^2 \ - \ 1120·x \ \ = \ \ 0 \ \ . $$ One of the roots is $ \ x \ = \ 0 \ \ , $ since we constructed it to have one intersection point of the curves on the $ \ y-$axis; the remaining cubic has a real zero at $ \ x \ \approx \ 1.186 \ $ and two complex-conjugate zeroes. [In regard to the earlier remark, we also have two intersections between the curves for $ \ c \ = \ 3 \ \ . ] $
The generic intersection equation is the quartic $$ 25·x^4 \ + \ (10c + 250)·x^3 \ + \ (c^2 + 100c + 500)·x^2 \ + \ (10c^2 + 220c - 1350)·x \ \ = \ \ 0 \ \ , $$ for which we again see that $ \ x \ = \ 0 \ $ is a root. We would really rather not have to solve for the real zero of the residual cubic polynomial. Indeed, we won't need to: we find that the linear term in the polynomial is equal to zero for $$ 10c^2 \ + \ 220c \ - \ 1350 \ \ = \ \ 0 \ \ \Rightarrow \ \ c \ \ = \ \ \frac{-220 \ \pm \ \sqrt{220^2 \ + \ 4·10·1350}}{2·10} \ \ = \ \ -11 \ \pm \ 16 \ \ . $$ For $ \ c \ = \ 5 \ \ , $ the intersection equation "collapses" to $$ 25·x^4 \ + \ 300·x^3 \ + \ 1025·x^2 \ \ = \ \ 25x^2 \ · \ ( \ x^2 \ + \ 12·x \ + \ 41 \ ) \ \ = \ \ 0 \ \ , $$ for which $ \ x \ = \ 0 \ $ is the only real root. The corresponding parabola is then $ \ y \ = \ x^2 + \left(5 + \frac{5}{5} \right)·x + 5 \ = \ x^2 + 6x + 5 \ \ . $
Our second solution, $ \ c \ = \ -27 \ \ $ gives us the parabola $ \ y \ = \ x^2 + \left(5 + \frac{-27}{5} \right)·x - 27 \ = \ x^2 - \frac25·x - 27 \ \ . $ However, this doesn't meet the conditions set forth in the problem statement, since our constructed circle meets the parabola on the $ \ y-$axis, close to its vertex at $ \ \left( \frac15 \ , \ -27.04 \right) \ \ $ , so $ \ (0 \ , \ -27) \ $ plainly cannot be the closest point on the parabola to $ \ (12 \ , \ 3) \ \ . $
Thus, $ \ \mathbf{y \ = \ x^2 + 6x + 5} \ $ is the only parabola with the stated properties.
On
$ y = x^2 + bx + c $
$ y' = 2 x + b $
At $(0,c)$ we have $y' = b $, and thus
$ b \bigg( \dfrac{c - 3}{0 - 12} \bigg) = -1 $
which simplifies to $ b (c - 3) = 12 $
We also have $(-5, 0)$ on the parabola, then $0 = 25 - 5 b + c $
Solving the above two equations yields two solutions
Case I: $b = -0.4, c = -27 $
Case II: $ b = 5 , c = 6 $
We need to check if the distance at the $y-intercept$ is indeed the shortest distance. At the critical points, we have
$ (2 x + b ) \bigg( \dfrac{x^2 + bx + c - 3 }{ x - 12 } \bigg) = -1 $
So that
$ 2 x^3 + 3 b x^2 + x ( 1 + 2 (c - 3) + b^2) + b ( c - 3) - 12 = 0 $
In case I, this gives three critical points, listed with their respective distances from $(12, 3)$
$A (-5.13231, 1.393538) , 17.20746 $
$B (0, -27) , 32.31099 $
$ C (5.732311, 3.566462), 6.293235 $
Clearly the $y$-intercept, which is point $B$ is not the closest point to $(12,3)$. Hence Case I is extraneous.
Case II leads to one critical point only which is $(0, 5)$ with a distance of $12.16553$.
Hence our parabola is
$ y =x^2 + 6 x + 5 $
Let $P$ be the parabola's $y$-intercept $(0,c)$ and $Q$ be $(12,3).$
Then the parabola's tangent at $P$ is perpendicular to $PQ,$ and, since $P$ and $Q$ have different $x$-coordinates, $PQ$ is not vertical. So, $$(2x+b)\left(\frac{c-3}{0-12}\right)=-1\quad\text{and}\quad x=0\\b(c-3)=12.$$ And as $c=5b-25,$ $$\\5b^2-28b-12=0\\b=6\quad\text{or}\quad -\frac25\\c=5\quad\text{or}\quad -27.$$
If $(b,c)=\left(-\frac25,-27\right),$ then the parabola is $\displaystyle y=x^2-\frac25x-27,$ and its points whose tangents are perpendicular to $PQ$ are given by $$\left(2x-\frac25\right)\left(\frac{x^2-\frac25x-27-3}{x-12}\right)=-1\\x=-5.13\quad\text{or}\quad 0 \quad\text{or}\quad 5.73,$$ and the corresponding distances from $(12,3)$ are approximately $17,32$ and $7.$ We eliminate this case since here the point on the parabola closest to $(12,3)$ is not the $y$-intercept.
SIMPLER ALTERNATIVE (suggested by Oscar Lanzi below): If $b=-\frac25,$ then the parabola's axis $x={-}\frac b2$ is $x=\frac15,$ and $P$ and $Q$ lie on opposite sides of it, which means that some point on the parabola is closer than $P$ to $Q;$ so, we eliminate this case.
Hence, the required parabola must be $y=x^2+6x+5.$