Area of a locus?

Before one attempts to solve this problem one must visualize this:

Let $U$ = union of all lines $L_1$ . Let $V$ = union of all lines $L_2$ . Every point in the region $U \cap V$ is a candidate intersection point of the lines $L_1$ and $L_2$ . Therefore, $L = U \cap V$ such that the intersection point lies inside the circle.

Some Facts: Further information

1. $y = mx + \dfrac{a}{m}$ is a tangent to the parabola $y^2 = 4ax$ , whatever be the value of $m$ .

2. $y = mx -am^2$ is a tangent to the parabola $x^2 = 4ay$ , whatever be the value of $m$ .

3. For any point $P$ , that is outside a parabola, there is at least one tangent on the parabola passing through $P$ .

Study the set of lines $L_1$ :

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: $\dfrac{-\dfrac{3a}{5}+b-(-2)}{-\dfrac{3}{5}-(3)} *a = -1$ $\implies b= \dfrac{3a^2-10a+18}{5a} = \dfrac{3a}{5} -2 + \dfrac{18}{5a}$ This means we can eliminate one variable and write the equation of $L_1: y= ax+ \dfrac{3a^2-10a+18}{5a}$ for all $a$ .

$L_1: y+2= a(x+ \dfrac{3}{5})+\dfrac{18}{5a}$ . Let $Y = y+2, X = x+ \dfrac{3}{5} \implies Y = aX + \dfrac{18}{5a}$ . Using fact 1, we get that $L_1$ is always tangent to parabola: $Y^2 = \dfrac{72X}{5}$ . Upon back substitution, we get: $(y+2)^2 = \dfrac{72(x+ \dfrac{3}{5})}{5} \implies 25(y+2)^2 = 72(5x+3)$ . So in conclusion the set of lines $L_1$ are just the set of all tangents to this parabola: ( $P_1$ ). Using fact 3, for any point that is outside $P_1$ , a tangent line from $P_1$ can be drawn to that point. Hence union of all tangents is the region outside $P_1$ .

Study the set of lines $L_2$ :

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: $\dfrac{-\dfrac{3}{5}+2}{\dfrac{-\dfrac{3}{5} -d}{c}-3} *c = -1$ $\implies d= \dfrac{7c^2-15c-3}{5} = \dfrac{7c^2}{5} - 5c - \dfrac{3}{5}$ This means we can eliminate one variable and write the equation of $L_2: y= cx+ \dfrac{7c^2-15c-3}{5}$ for all $c$ .

$L_2: y+ \dfrac{3}{5}= c(x- 3)+\dfrac{7c^2}{5}$ . Let $Y = y+ \dfrac{3}{5}, X = x- 3\implies Y = cX + \dfrac{7c^2}{5}$ . Using fact 2, we get that $L_2$ is always tangent to parabola: $X^2 = -\dfrac{28Y}{5}$ . Upon back substitution, we get: $(X-3)^2 = -\dfrac{28(y+ \dfrac{3}{5})}{5} \implies y= \dfrac{-5(x-3)^2}{28} - \dfrac{3}{5} (P_2)$ .

So overall, for any point outside both $P_1$ and $P_2$ a tangent can be drawn to both $P_1$ and $P_2$ from that point, so the locus of intersection points of $L_1$ and $L_2$ is any point outside both the parabolas. But we must remember that the intersection point must be inside the circle because we are looking for the locus of the intersection of the chords $PQ$ and $RS$ , both of which must reside inside the circle, hence their intersection will also be inside the circle. So the region is: $L: \{(x,y) \in \mathbb{R}: y> \dfrac{-5(x-3)^2}{28} - \dfrac{3}{5}, 25(y+2)^2 > 72(5x+3), (x-3)^2 + (y+2)^2<25 \}$

To make finding the area simpler lets shift this region (make the circle have center at origin): $X = x+3, Y = y-2$ $\implies L: \{(X,Y) \in \mathbb{R}: Y> \dfrac{-5(X)^2}{28} - \dfrac{3}{5}, 25(y)^2 > 72(5x+18), (x)^2 + (y)^2<25 \}$

$A: \left(- \dfrac{11}{5}, \dfrac{6\sqrt{14}}{5}\right); B: \left(-\dfrac{2(6\sqrt{7}-7)}{5}, - \dfrac{12(\sqrt{7}-3)}{5}\right); C: \left(-\dfrac{6\sqrt{14}}{5}, - \dfrac{11}{5}\right)$ .

The longest distance $d$ between any two points in the region is obviously AC because A is the point with both maximum x and y coordinates. Any point on $P_2$ (in the shaded region) is closer to A than C because both the x and y coordinates is closer to A. Also any other point along the circle is a chord and C is farthest away as none of the points is the diameter, hence C is the farthest point from A. $d = AC = \sqrt{50} = 5\sqrt{2}$ .

Finding the area is a tedious task to do by hand. It can be computed as follows:

Region above the axis: $\int_{-5}^{-\frac{18}{5}}\sqrt{25-x^{2}}dx+\int_{-\frac{18}{5}}^{-\frac{11}{5}}\left(\sqrt{25-x^{2}} - \sqrt{\frac{72\left(5x+18\right)}{25}} \right)dx$

Region below the axis: $\left|\int_{-5}^{-\frac{6\sqrt{14}}{5}}-\sqrt{25-x^{2}}dx+\int_{-\frac{6\sqrt{14}}{5}}^{-\frac{18}{5}}\left(-\frac{5}{28}\left(x\right)^{2}+\frac{7}{5}\right)dx+\int_{-\frac{18}{5}}^{-\frac{2\left(6\sqrt{7}-7\right)}{5}}\left(\left(-\frac{5}{28}\left(x\right)^{2}+\frac{7}{5}\right)-\left(-\sqrt{\frac{72\left(5x+18\right)}{25}}\right)\right)dx\right|$

Adding both these values we get area of region $= 6.94701 \implies \lfloor{1000*6.94701}\rfloor = \lfloor{6947.01...}\rfloor = 6947$