Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$

Let $x=\cos\theta$ and $y=\sin\theta$,because $\sin^2\theta+\cos^2\theta=1$.
Then we need to find the minimum and maximum value of the expression $\frac{4-\sin\theta}{7-\cos\theta}$.
I differentiated it and equated it to zero to find the critical points or points of extrema.
They are $\theta_1=\arcsin(\frac{1}{\sqrt{65}})-\arctan(\frac{7}{4})$ and $\theta_2=\arccos(\frac{1}{\sqrt{65}})+\arctan(\frac{4}{7})$
I found $\frac{4-\sin\theta_1}{7-\cos\theta_1}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}$.
$\frac{4-\sin\theta_1}{7-\cos\theta_1}=\frac{3}{4}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}=\frac{5}{12}$
This method is full of lengthy calculations.I want to know is there an elegant solution possible for this problem which is short and easy.

Let $\displaystyle k=\frac{4-y}{7-x}\Rightarrow 7k-kx=4-y\Rightarrow kx-y = 7k-4$ and given $x^2+y^2=1$
Now using the Cauchy-Schwarz inequality, we get $$[k^2+(-1)^2](x^2+y^2)\geq (kx-y)^2$$
So $$k^2+1\geq (7k-4)^2\Rightarrow 49k^2+16-56k\leq k^2+1$$
So $$48k^2-56k+15\leq 0\Rightarrow (4k-3)(12k-5)\leq 0$$
So we get $\displaystyle \frac{5}{12}\leq k\leq \frac{3}{4}$
