Area of $(x-3)^2+(y+2)^2<25: (x,y) \in L_1 \cap L_2$ Two lines $(L_1,L_2)$ intersects the circle $(x-3)^2+(y+2)^2=25$ at the points $(P,Q)$ and $(R,S)$ respectively. The midpoint of the line segment $PQ$ has $x$-coordinate $-\dfrac{3}{5}$, and the midpoint of the line segment $RS$ has $y$-coordinate $-\dfrac{3}{5}$.
If $A$ is the locus of the intersections of the line segments $PQ$ and $RS$, then the area of the region $A$ is:

What I've done:
Consider $L_1: y= ax+b$. The midpoint of the chord $PQ$ is $(-\dfrac{3}{5}, -\dfrac{3a}{5}+b)$. Now, using the property that the midpoint of a chord of a circle and the center of the circle $(3,-2)$ are perpendicular we have: $\dfrac{-\dfrac{3a}{5}+b-(-2)}{-\dfrac{3}{5}-(3)} *a = -1$
$$\implies b= \dfrac{3a^2-10a+18}{5a}$$ This means we can eliminate one variable and write the equation of $L_1: y= ax+ \dfrac{3a^2-10a+18}{5a}$
From this form of $L_1$ we can get the value of the minimum value of the $y$-intercept by differentiating $b$. Let $b= f(a) = \dfrac{3a^2-10a+18}{5a}, f'(a) = \dfrac{3a^2-18}{5a^2}, f'(a)=0 \implies a = \pm \sqrt{6}$. I just found this hoping we will get bounds on the y-intercept of $L_1$.
Now lets do the same process for $L_2$.
Consider $L_2: y= cx+d$. The midpoint of the chord $RS$ is $(\dfrac{-\dfrac{3}{5} -d}{c},-\dfrac{3}{5})$. Now, using the property that the midpoint of a chord of a circle and the center of the circle $(3,-2)$ are perpendicular we have: $\dfrac{-\dfrac{3}{5}+2}{\dfrac{-\dfrac{3}{5} -d}{c}-3} *c = -1$
$$\implies d= \dfrac{7c^2-15c-3}{5}$$ This means we can eliminate one variable and write the equation of $L_2: y= cx+ \dfrac{7c^2-15c-3}{5}$
From this form of $L_2$ we can get the value of the minimum value of the $y$-intercept by differentiating $d$. Let $d= f(c) = \dfrac{7c^2-15c-3}{5}, f'(c) = \dfrac{14c-15}{5}, f'(c)=0 \implies c = \dfrac{15}{14}$. I just found this hoping we will get bounds on the y-intercept of $L_2$.
Along with all this, we can set bounds when the line segment is just about to leave the circle (tangent to the circle).
What I can visualize:
Let $X$ = union of all line segments $PQ$.
Let $Y$ = union of all line segments $RS$.
Every point in the intersection of $X$ and $Y$ is a candidate intersection point of the lines $L_1$ and $L_2$. So A = $X \cap Y$
Edit 1: I saw the equation of $L_1$ varying as $a$ varies on DESMOS and think the boundry of the union of all line segments PQ might be an outer circle.

EDIT (Original answer at the end).
I want to show how the envelope of chords $RS$ (or $PQ$) can be obtained without calculus and without coordinates (see figure below).
Let's start with a chord $AB$ of a circle of centre $O$. For any point $M$ on that chord, we can construct a line $RS$ passing through $M$ and perpendicular to $OM$. We want to find the envelope of all those lines, i.e. the curve which is tangent to all the lines $RS$ as $M$ varies on $AB$.
Consider then another point $M'$ on $AB$ and its associated line $R'S'$. Let $P$ be the intersection of $RS$ and $R'S'$, $T$ the tangency point of $RS$ with the envelope, $T'$ the tangency point of $R'S'$ with the envelope. As $M'$ approaches $M$, both $T'$ and $P$ approach $T$.
But the circle through $OMM'$ also passes through $P$ (because $\angle PMO=\angle PM'O=90°$) and this circle, as $M'\to M$, tends to the circle through $O$ tangent to $AB$ at $M$. Hence $T$, which is the limiting position of $P$, is the intersection of that circle with line $RS$. Moreover, $OT$ is a diameter of that circle.

Now that we know how to construct point $T$ on $RS$,
let's also construct line $HK$, parallel to $AB$ at a distance from it equal to the distance of $O$ from $AB$. If $J$ is the projection of $T$ on it, then $TJ=TO$, because line $CH$ joining the midpoints of the legs of a trapezoid is the arithmetic mean of bases $OK$ and $TJ$.
It follows that point $T$ has the same distance from $O$ and from line $HK$. Hence its locus (which is the envelope) is a parabola, having $O$ as focus and $HK$ as directrix.

ORIGINAL ANSWER.
What you need is the envelope $\gamma_1$ formed by lines $L_1$ and the envelope $\gamma_2$ formed by lines $L_2$.
As the equation of $L_1$ is $y=\left(x+{3\over5}\right)a-2+{18\over5a}$, differentiating w.r.t. $a$ we get: $x+{3\over5}-{18\over5a^2}=0$, which can be solved for $a$:
$$
a^2={18\over 5x+3}.
$$
Inserting this into the equation of $L_1$ we get (after some algebra):
$$
25(2+y)^2={72}(5x+3),
$$
which is the desired envelope $\gamma_1$ (a parabola).
Repeating the same process for $L_2$ we can find the equation of $\gamma_2$ (another parabola):
$$
y=-{5\over28}(3-x)^2-{3\over5}.
$$
Lines $L_1$ and $L_2$ are tangent to their envelope, hence
the area you want to compute is that external to both parabolas but inside the circle.

To compute the area, one has to find the coordinates of the upper intersection $A$ between $\gamma_1$ and the circle,
of the left intersection $C$ between $\gamma_2$ and the circle
and of the intersection $B$ between $\gamma_1$ and $\gamma_2$
lying inside the circle:
$$
A=\left(\frac{4}{5},\frac{6}{5} \sqrt{14}-2\right),\quad
C=\left(3-\frac{6}{5}\sqrt{14},-\frac{21}{5}\right),\quad
B=\left(\frac{1}{5} \left(29-12 \sqrt{7}\right),\frac{2}{5} \left(6
   \sqrt{7}-23\right)\right).
$$
Integrating along $y$ the area can then be computed as:
$$
\begin{align}
area&=\int_{y_B}^{y_A} 
\left[\left(\frac{5}{72}(y+2)^2-\frac{3}{5}\right)-\left(3-\sqrt{25-(y+2)^2}\right)\right] \, dy \\
&+\int_{y_C}^{y_B}
\left[\left(-\frac{2}{5} \sqrt{-35 y-21}+3\right)-\left(3-\sqrt{25-(y+2)^2}\right)\right] \, dy \\
&=\frac{25 \pi }{4}+4 \sqrt{14 \left(9-4 \sqrt{2}\right)}-\frac{3004}{75}.
\end{align}
$$
