Implicit Geometry

We're given that $(p_1 - v_1)^T p_1 = 0$ , therefore, $p_1^T p_1 - p_1^T v_1 = 0$ . This is can re-written as

$(p_1 - \dfrac{v_1}{ 2})^T (p_1 - \dfrac{v_1}{ 2}) = (\dfrac{v_1}{ 2})^T(\dfrac{v_1}{ 2})$

And this is an equation of a circle with center at $\dfrac{v_1}{ 2} = (0, \dfrac{1}{2})$ , and radius $r_1 = \dfrac{1}{2}$ .

The same can be said about the second curve, resulting in $p_2$ being a circle with center at $(1, 0)$ and radius $r_2 = 1$ . Now comes the fun part. The distance between the two centers $= \sqrt{ 1 + 1/4} = \dfrac{\sqrt{5}}{2}$ while the two radii add to 1.5, which means there is an intersection between the two circles. The two centers with an intersection point form a triangle with sides $\frac{1}{2} , 1 , \frac{\sqrt{5}}{2}$ . This makes it possible to determine the angles subtended by the intersecting arcs at the two centers. In what follows, $\theta_1$ and $\theta_2$ are half of the subtended angles. It is easy to see that,

$\theta_1 = \cos^{-1} ( \dfrac{\frac{5}{4} + \frac{1}{4} - 1}{ 2 \cdot \frac{1}{2} \cdot \frac{ \sqrt{5}}{2}} )= \cos^{-1} ( \dfrac{1}{ \sqrt{5} })$

while,

$\theta_2 = \cos^{-1} ( \dfrac{ \frac{5}{4} + 1 - \frac{1}{4}}{ 2 \cdot 1 \cdot \frac{\sqrt{5}}{2}} ) = \cos^{-1} ( \dfrac{2}{ \sqrt{5} })$

Now we're ready to find the area.

$\text{Area} = \frac{1}{2} r_1^2 (2 \theta_1) + \frac{1}{2} r_2^2 (2 \theta_2) - \frac{1}{2} r_1^2 \sin(2 \theta_1) - \frac{1}{2} r_2^2 \sin(2 \theta_2)$

Plugging in all the values results in,

$\text{Area} = \boxed{0.2404348}$