If real $a$ and $b$ satisfy $17(a^2+b^2)-30ab-16=0$, then find the maximum value of $\sqrt{16a^2+4b^2-16ab-12a+6b+9}$ If $ a $ and $ b $ are real numbers that satisfy the equation $ 17 (a ^ 2 + b ^ 2) -30ab-16 = 0 $, the maximum value of the expression$$\sqrt{16a^2+4b^2-16ab-12a+6b+9}$$goes:
a) $1$
b) $3$
c) $5$
d) $7$
e) $9$
Can anyone give a hint?
Attempt:  I tried Variation, Jacobian, Hessian matrices.

The hint.
Let $2a-b=k.$
Thus, we need to find a maximal value of $\sqrt{4k^2-6k+9}.$
Now, find all values of $k$, for which the equation 
$$k^2(17(a^2+b^2)-30ab)=16(2a-b)^2$$ has solutions and choose from these values such that $\sqrt{4k^2-6k+9}$ will get a maximal value. 
I got that it happens for $k=-\frac{5}{2}.$
