Find common points of $(x^2+y^2)^2=8xy $ and $(x-1)^2+(y-1)^2=1$

Question

Im trying to count an area limited by these two curves, but one step of my solution needs to find their intersection points, so i can find angle which i have to put to polar coordinates. When i start to count it in normal way i encounter really horrible equations and since $(x-1)^2+(y-1)^2=1$ doesn't have center at $(0,0)$ as this lemniscate has, i can't change to polar coordinates. Any idea how can i deal with it?

Answer 1

Expanding $(x-1)^2+(y-1)^2 = 1$, we get $x^2 + y^2 = 2x + 2y-1$. Substituting this into the other equation, we thus obtain \begin{equation*}(2x+2y-1)^2=8xy\end{equation*} When we expand the left-hand side, there is an $8xy$ term which cancels with that on the right-hand side, leaving \begin{equation*}4x^2+4y^2-4x-4y+1=0\end{equation*} Now we again substitute $x^2+y^2=2x+2y-1$, giving \begin{equation*}4(2x+2y-1)-4x-4y+1=0 \iff 4x+4y-3=0 \iff y = \frac{3}{4}-x\end{equation*} Substituting this into $(x-1)^2+(y-1)^2=1$, we deduce \begin{align*}(x-1)^2+\left(\frac{3}{4}-x-1\right)^2 = 1 &\iff 32x^2-24x+1=0 \\ &\iff x = \frac{3\pm\sqrt{7}}{8} \\ &\iff y = \frac{3 \mp \sqrt{7}}{8}\end{align*} so the common points are $\left(\frac{3+\sqrt{7}}{8},\frac{3-\sqrt{7}}{8}\right)$ and $\left(\frac{3-\sqrt{7}}{8},\frac{3+\sqrt{7}}{8}\right)$.

Answer 2

Perhaps this figure will help you visualize the solution:

Answer 3

In polar coordinates, $x = r\cos\theta, y = r\sin\theta, x^2+y^2 = r^2$

$r^4 = 8r^2\cos\theta\sin\theta\\ r^2 = 4\sin 2\theta\\ r = 2\sqrt{\sin 2\theta}$

and

$r^2 - 2r\cos\theta - 2r\sin\theta = - 1\\ r^2 - 2r\cos\theta - 2r\sin\theta + (\cos\theta + \sin\theta)^2 = (\cos\theta + \sin\theta)^2- 1\\ (r - (\cos\theta + \sin\theta))^2 = \sin2\theta\\ r = \sqrt{\sin 2\theta} + (\cos\theta + \sin\theta)$

$2\sqrt {\sin 2\theta} = |\sqrt{\sin 2\theta} + (\cos\theta + \sin\theta)|$

$\sqrt {\sin 2\theta} = \cos\theta + \sin\theta\\ \sin 2\theta = 1 + \sin 2\theta$

or

$3\sqrt{\sin 2\theta}= (\cos\theta + \sin\theta)\\ 9\sin 2\theta = 1 + \sin 2\theta\\ \sin 2\theta = \frac {1}{8}\\ \theta = \frac 12 \arcsin \frac 18\\ \theta = \frac 12 (\pi - \arcsin \frac 18)\\ $

$r = \frac{\sqrt 2}{2}$