A familiar curve!

Find $q$ with respect to $p$ : $\text{Solve}[\text{EuclideanDistance}[\{p,0\},\{0,q\}]=1,q]$ is $q = \sqrt{1-p^2}$ .

Draw a picture using various lines connecting the $\{p,0\}$ and $\left\{0,\sqrt{1-p^2}\right\}$ points with a quarter circle for comparison:

Now, it is known that the arclength is less than that of a quarter circle with a radius of 1.

Following the procedure for computing the envelope of a family of curves and then computing the arc-length of that curve gives the desired answer, which is $\frac{3}{2}$ .

The members of the family of curves (straight lines in this case) are: $y=\sqrt{1-p^2}-\frac{\sqrt{1-p^2} x}{p}$ .

$\frac{\partial (\text{yn}-y)}{\partial p}\text{ is }-\frac{\sqrt{1-p^2} x}{p^2}-\frac{x}{\sqrt{1-p^2}}+\frac{p}{\sqrt{1-p^2}}$ .

Solving for p $-\frac{\sqrt{1-p^2} x}{p^2}-\frac{x}{\sqrt{1-p^2}}+\frac{p}{\sqrt{1-p^2}}=0$ gives three roots, but only one real root: $\sqrt[3]{x}$

Substituting that value for $p$ into the original family formula gives the curve formula: $y=\left(1-x^{2/3}\right)^{3/2}$ .

Drawing both the original family of straight lines and the curve as a broad, red curve shows that it is probably correct.

$\int \sqrt{\left(\frac{\partial \left(1-x^{2/3}\right)^{3/2}}{\partial x}\right)^2+\left(\frac{\partial x}{\partial x}\right)^2} \, dx$ gives $\frac{3 x^{\frac{3}{2}}}{2}$ .

$\left|\frac{3 x^{\frac{3}{2}}}{2}\right|_{x=0}^{1}=\frac{3}{2}$

From there, finding the arc length is pretty standard.

The shape is an Astroid , to see why you may want to check out this problem and it's solutions. An astroid has parametrization $(\cos^3{t},\sin^3{t})$ , so it's length in the first quadrant is $\large \int_0^{\frac{\pi}{2}}\sqrt{\left(\frac{d\sin^3t}{dt}\right)^2+\left(\frac{d\cos^3t}{dt}\right)^2}dt=\int_0^{\frac{\pi}{2}}\sqrt{\left(3\left(\sin^2t\right)\cos t\right)^2+\left(3\left(\cos^2t\right)\sin t\right)^2}dt$

Which simplifiest to $\large \frac{3}{2}\int_0^{\frac{\pi}{2}}\sin\left(2t\right)dt=1.5$