A geometry problem by Priyanshu Mishra

$\begin{aligned} x^2+y^2+14x+6y-6 & = 0 \\ (x+7)^2 - 49 + (y+3)^2 - 9 - 6 & = 0 \\ (x+7)^2 + (y+3)^2 & = 64 \end{aligned}$

That is an equation of a circle with a radius pf 8. We can let $\begin{cases} x+7 = 8\cos \theta & \implies x = 8\cos \theta - 7 \\ y+3 = 8\sin \theta & \implies y = 8\sin \theta - 3 \end{cases}$

Therefore,

$\begin{aligned} X & = 3x + 4y \\ & = 3(8\cos \theta - 7)+4(8\sin \theta - 3) \\ & = 24\cos \theta + 32\sin \theta - 33 \\ & = 40\left(\frac 35 \cos \theta + \frac 45 \sin \theta \right) - 33 \\ & = 40\sin \left(\theta + \tan^{-1} \frac 34 \right) - 33 & \small {\color{#3D99F6} X \text{ is maximum when }\sin \left(\theta + \tan^{-1} \frac 34 \right)=1} \\ \implies X_{max} & = 40-33 = \boxed{7} \end{aligned}$