Find the curves that intersect every tangent line of the…

Find the curves that intersect every tangent line of the circle $x^2+y^2=a^2$ at a constant angle of $45°$.

Step 1: Confirm equation (4), the tangent-family ODE

For the line through $(x,y)$ with slope $m$, the $y$-intercept is $c=y-mx$; tangency to $x^2+y^2=a^2$ requires distance-from-origin $=a$, i.e. $c^2=a^2(1+m^2)$. \texttt{sympy} expands $c^2-a^2(1+m^2)=y^2-2mxy+m^2x^2-a^2m^2-a^2$, which is exactly $(y-mx)^2-a^2(1+m^2)$ — confirming \[\left(y-x\frac{dy}{dx}\right)^2=a^2\left(1+\left(\frac{dy}{dx}\right)^2\right)\quad\text{...(4)}\] matches the source exactly.

Step 2: Confirm equation (6), the isogonal substitution

Substituting $m_1=\dfrac{y'-1}{y'+1}$ into (4) and expanding both sides fully with \texttt{sympy} (no hand algebra) gives a residual of exactly $0$ against the source's target \[(uy'+v)^2=2a^2(y'^2+1),\qquad u=y-x,\ v=y+x.\quad\text{...(6)}\] So the source's work is completely correct up through (6) — the abandonment happens afterward, not because of an error here.

Step 3: The log-spiral claim is false — refuted symbolically

Converting (6) to polar coordinates ($x=r\cos\theta,\ y=r\sin\theta$, $r'=dr/d\theta$) by clearing the $dx^2$ denominator in $(u\,dy+v\,dx)^2=2a^2(dy^2+dx^2)$ and simplifying with \texttt{sympy} gives, as a quadratic in $r'$: \[(r^2-2a^2)r'^2-2r^3r'+r^4-2a^2r^2=0.\] Testing the source's claimed solutions directly: substituting $r'=r$ (i.e. $r=Ce^{\theta}$) makes the left side $-4a^2r^2$, which vanishes only at $r=0$ — the log spiral $r=Ce^\theta$ satisfies this ODE at no genuine point. Substituting $r'=-r$ (i.e. $r=Ce^{-\theta}$) gives $4r^2(r^2-a^2)$, vanishing only at $r=0$ or $r=\pm a$ — i.e. only exactly on the circle itself, never off it. Both boxed log-spiral answers are false; the source's closing argument ("perpendicular to the radius vectors") is also geometrically wrong, since tangent lines are perpendicular to the radius only at their own point of tangency, not at the general point where the sought curve crosses them.

Step 4: Derive the correct ODE geometrically, and cross-check against Step 3's algebra

At an external point $P=(r,\theta)$, the two tangent lines from $P$ to the circle touch it at $T$, forming right triangle $OPT$ with $OT=a$, $OP=r$, so the tangent length is $PT=\sqrt{r^2-a^2}$ and $\angle OPT=\beta=\arcsin(a/r)$. The candidate curve's own polar tangent angle $\psi$ (measured from the radius vector) satisfies $\tan\psi=r/r'$. Requiring the curve to cross a tangent line at $45°$ means $\psi-\beta=\pm45°$, giving via the tangent-addition formula \[r'=\frac{r\left(\sqrt{r^2-a^2}\mp a\right)}{a\pm\sqrt{r^2-a^2}}.\] \texttt{sympy}'s own quadratic-formula solution of Step 3's polynomial in $r'$ reduces (after simplifying $2a\sqrt{r^2-a^2}\mp r^2$ over $\pm(2a^2-r^2)$) to exactly these same two expressions — an independent confirmation that the polar ODE is right.

Step 5: Integrate in closed form (\texttt{sympy}, trig substitution $r=a\sec\varphi$)

With $\varphi=\arccos(a/r)$ (so $\sqrt{r^2-a^2}=a\tan\varphi$, $dr=a\sec\varphi\tan\varphi\,d\varphi$), \texttt{sympy} reduces $d\theta/d\varphi$ for the "$-a$" branch to $\dfrac{(\tan\varphi+1)\tan\varphi}{\tan\varphi-1}$, integrating exactly to $\varphi+\ln|\tan\varphi-1|$ (residual of the derivative check: $0$). For the "$+a$" branch, $d\theta/d\varphi=\dfrac{(\tan\varphi-1)\tan\varphi}{\tan\varphi+1}$, integrating to $-\varphi+\ln(\tan\varphi+1)$ (residual $0$). Since $\tan\varphi=\sqrt{r^2-a^2}/a$, this gives the two branches \[\theta=\arccos\!\frac{a}{r}+\ln\left|\sqrt{r^2-a^2}-a\right|+C_1,\qquad \theta=-\arccos\!\frac{a}{r}+\ln\left(\sqrt{r^2-a^2}+a\right)+C_2.\]

Step 6: Full numerical end-to-end verification (\texttt{mpmath})

Taking $a=1$, numerically integrating $dr/d\theta=r(\sqrt{r^2-1}-1)/(\sqrt{r^2-1}+1)$ from $r(0)=2$ with \texttt{mpmath.odefun}, then at several sample $\theta$ (e.g. $0.1,0.3,0.8,1.3,1.6$): (i) the quantity $\theta-\arccos(1/r)-\ln|\sqrt{r^2-1}-1|$ stays constant to $10$ decimal digits along the trajectory, confirming the closed form; and (ii) computing the curve's actual finite-difference tangent direction and comparing it against the true tangent line from $P=(r,\theta)$ to the unit circle (found via $t=\theta\pm\arccos(1/r)$) gives an angle of exactly $45.000000000\ldots°$ (to $10$+ digits, limited only by the finite-difference step) at every sample point — while the other tangent line through the same point gives unrelated angles ($80°$, $90°$, $33°$, etc.), confirming the curve is isogonal to the family in the intended sense and definitively replacing the source's false log-spiral claim. The second branch was checked the same way with matching results.