Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian

Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$

$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
 A: Since
$$
x^2+Ax+\left( \frac{A}{2} \right) ^2+y^2+By+\left( \frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C\\
\left( x+\frac{A}{2} \right) ^2+\left( y+\frac{B}{2} \right) ^2=\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C
$$
Let $\displaystyle r=\sqrt{\left( \frac{A}{2} \right) ^2+\left( \frac{B}{2} \right) ^2-C}$, we can set $x=r\cos \theta -\dfrac{A}{2},y=r\sin \theta -\dfrac{B}{2}$, then
$$
\begin{aligned}
7x-24y&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} (7 \cos \theta-24 \sin \theta)-7 A+24 B\right)\\
&=\frac{1}{2} \left(\sqrt{A^2+B^2-4 C} \sqrt{7^2+24^2}\sin(\theta+\phi)-7 A+24 B\right)\\
&\geq \frac{1}{2} \left(25\sqrt{A^2+B^2-4 C} -7 A+24 B\right)
\end{aligned}
$$
The second step uses The auxiliary Angle formula.
A: Using Cauchy-Schwarz inequality,
$$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\
\implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\
\ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\right)^2}+ \frac{-7A+24B}{2}\\
=-25 \frac{1}{2} \sqrt{A^2+B^2-4C} + \frac{-7A+24B}{2}\\
=\frac 12 \left(-25\sqrt{A^2+B^2-4C}-7A+24B\right).\blacksquare$$
A: Hint:
Let $7x-24y=z\iff x=?$
Replace the value of $x$ in $$x^2+y^2+Ax+By+C=0$$ to form a Quadratic Equation in $y$
As $y$ is real, the discriminant must be $\ge0$
