I will assume that the circle has radius equal to $3$. If that is the case, then the circle is defined by
$C := \{ (x,y) \in \mathbb{R}^2 \mid (x-8)^2 + (y-7)^2 = 3^2\}$.
The equation that defines the circle can be rewritten in the form
$(y-7)^2 = 9 - (x-8)^2$
and, taking the square root of both sides, we obtain
$y = 7 \pm \sqrt{9 - (x-8)^2}$.
Since we want the top half of the circle, we have $y = 7 + \sqrt{9 - (x-8)^2}$.
The line that connects points $(1,-3)$ and $(5,7)$ is given by the equation
$y + 3 = \frac{10}{4} (x-1)$
or, equivalently,
$y = \displaystyle\frac{5}{2} x - \frac{11}{2}$.
Finally, we can introduce a function $f : [1,11] \to \mathbb{R}$ defined by
$f (x) = \left\{\begin{array}{cl} \displaystyle\frac{5}{2} x - \frac{11}{2}, &\quad{} x \in [1,5]\\ 7 + \sqrt{9 - (x-8)^2}, &\quad{} x \in [5,11]\end{array}\right.$