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Given that x^2+y^2 = 14x+6y+6, what is the largest possible value that 3x+4y can have?

how to solve it?? Give full solution..

Notice that this is an equation of a circle.
We can manipulate this equation like this:
$x^2-14x+y^2-6y=6$
$\Rightarrow x^2-2x\cdot7+7^2+y^2-2y\cdot3+3^2=6+7^2+3^2$
$\Rightarrow (x-7)^2+(y-3)^2=8^2$
By Cauchy–Schwarz inequality,
$\left(3(x-7)+4(y-3)\right)^2\leq(3^2+4^2)\left((x-7)^2+(y-3)^2\right)$
$\Rightarrow3x+4y-33\leq40$
$\therefore3x+4y\leq73$