Algebraic manipulation is an art

In this diagram, we see that for any point $(a,b)$ on the circle, we have $a^2 + b^2 = 1$ and the slope of the line from $(a,b)$ to the point $C$ is exactly $\frac{\beta - b}{\alpha - a} = f(a,b)$ . Therefore, we need to find the points on the circle which maximize and minimize the slope. Say this happens at the points $A$ and $B$ respectively. Hence $\gamma$ is the slope of $BC$ and $\delta$ is the slope of $AC$ .

Now this means that $AC$ and $BC$ are tangent to the circle. Hence $\angle OAC = \angle OBC = 90^{o}$ . We also know that $OA = OB = 1$ and $OC = \sqrt{\alpha^2 + \beta^2}$ . Furthermore let $\varepsilon = \angle OCA = \angle OCB$ and $\theta$ be the angle between $OC$ and the x-axis.

Then we have $\gamma = tan(\theta - \varepsilon)$ and $\delta = tan(\theta + \varepsilon)$ .

Using trigonometry, we find that $(1)$ $tan(\theta) = \frac{\beta}{\alpha}$ and $(2)$ $tan(\varepsilon) = \frac{1}{\sqrt{\alpha^2 + \beta^2 - 1}}$ .

The rest of my solution is algebra. It requires the following trigonometric identity: $tan(\theta + \varepsilon)tan(\theta - \varepsilon) = \frac{tan^2(\theta) - tan^2(\varepsilon)}{1 - tan^2(\theta)tan^2(\varepsilon)} . . . (3)$ Using $(1)$ , $(2)$ , $(3)$ and a bit of algebra, we get the following: $\gamma \times \delta = \frac{\beta^2 - 1}{\alpha^2 - 1} . . . . (4)$ Now we can use $(1)$ to get: $tan(2\theta) = \frac{2\alpha \beta}{\alpha^2 - \beta^2} . . . . (5)$ Now we also need the identity that $tan(\theta + \varepsilon) + tan(\theta - \varepsilon) = tan(2\theta)(1 - tan(\theta + \varepsilon)tan(\theta - \varepsilon)) . . . (6)$

Now we can use $(4)$ , $(5)$ and $(6)$ to get: $\gamma + \delta = \frac{2\alpha \beta}{\alpha^2 - 1} . . .(7)$ Next using the fact that $\gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma\delta$ , we get $\gamma^2 + \delta^2 = \frac{4\alpha^2 \beta^2 - 2(\alpha^2 - 1)(\beta^2 - 1)}{(\alpha^2 - 1)^2} . . .(8)$ Finally combine $(7)$ and $(8)$ : $\gamma + \delta + \sqrt{\gamma^2 + \delta^2} = \frac{2\alpha \beta + \sqrt{4\alpha^2 \beta^2 - 2(\alpha^2 - 1)(\beta^2 - 1)}}{\alpha^2 - 1}$ Hence $K+L+M = \boxed{8}$ .

See This Diagram image

$f(a,b)$ maximum and minimum values are geometrically means Slope of two tangents to the circle ${ x }^{ 2 }\quad +\quad { y }^{ 2 }\quad =\quad 1$ . from Point P( $\alpha$ , $\beta$ )

$\gamma \quad =\quad { m }_{ PB }\quad \quad and\quad \delta \quad =\quad { m }_{ PA }\quad$ .

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Now Try to proceed further independently So that You Realise Beauty of this question :)

Using polar coordinates $a = \cos\phi$ and $b = \sin\phi$ , function $f$ can be rewritten as: $f = \frac{\beta - \sin\phi}{\alpha - \cos\phi}\;.$ The extremum condition for $f$ , i.e. $f' = 0$ , reads: $\frac{\beta - \sin\phi}{\alpha - \cos\phi} = -\frac{1}{\tan\phi}\;.$ Note that $\phi \neq 0$ , since $\alpha > 1$ by hypothesis. Notice also that the above equation can be interpreted as the orthogonality condition between the line joining $(a,b)$ to $(\alpha,\beta)$ and the radius. Therefore, the former is tangent to the circle. The above equation can be solved for the sine and the cosine, $\sin\phi = \frac{\beta \pm \alpha\sqrt{\alpha^2 + \beta^2 - 1}}{\alpha^2 + \beta^2}\;,$ and $\cos\phi = \frac{\alpha \mp \beta\sqrt{\alpha^2 + \beta^2 - 1}}{\alpha^2 + \beta^2}\;.$ Thus, substituting back into $f$ , we get: $\gamma = \frac{\beta\sqrt{\alpha^2 + \beta^2 - 1} - \alpha}{\beta + \alpha\sqrt{\alpha^2 + \beta^2 - 1}}\;,$ and $\delta = -\frac{\beta\sqrt{\alpha^2 + \beta^2 - 1} + \alpha}{\beta - \alpha\sqrt{\alpha^2 + \beta^2 - 1}}\;.$ Summation gives: $\gamma + \delta = \frac{2\alpha\beta}{\alpha^2 - 1}\;,$ and multiplication gives: $\gamma\delta = \frac{\beta^2 - 1}{\alpha^2 - 1}\;.$ Therefore, $\gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma\delta = \frac{4\alpha^2\beta^2 - 2(\beta^2-1)(\alpha^2 -1)}{(\alpha^2 - 1)^2}\;.$ Finally, we can read off from the expression for $P$ that $K = 2$ , $L = 4$ and $M = 2$ , so that $\boxed{K + L + M = 8}\;.$

Interesting question deepanshu bhaiya. Solved in the similar way as you did.