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.

$$y-4=z(x-7)\tag{1}$$
is an equation of all the lines that pass $(7,4)$ with slope $z$. When $z$ is at its minimum or maximum then the line touches the unit circle. Equation of tangent lines with slope $z$ to a unit circle is
$$y=zx\pm\sqrt{z^2+1}\tag{2}$$
Comparing $(1)$ and $(2)$,
$$-7z+4=\pm\sqrt{z^2+1}$$
$$(7z-4)^2=z^2+1$$
$$48z^2-56z+15=0$$
$$(4z-3)(12z-5)=0$$
$$\therefore M=\frac34, m=\frac5{12}$$
