<s>[INST] Can two circles intersect each other at right angles, such that one circle passes through the center of the other circle? More precisely, do there exist two intersecting circles such that the tangent lines at the intersection points make an angle of 90 degrees, and one of the circles passes through the center of the other? I think the answer is no, and that this can only be the case if the circle passing through the center of the other circle is actually a line. I'm not sure how to show this though. Can someone give a proof or a counterexample?
 [/INST] Let
\begin{align}
\vec{r}_1(\alpha)&=R\left(\cos \alpha, \sin \alpha\right)\\
\vec{r}_2(\beta)&=\lambda R\left(1+\cos \beta, \sin \beta\right)
\end{align}
be equations of the given circles where $\lambda$ and $R$ are both positive. The parameters $\alpha$ and $\beta$ span from $0$ to $2\pi$.
Assume their intersections occur at $\alpha = a$ and $\beta = b$, we have
\begin{cases}
R\cos a = \lambda R \left(\cos b + 1\right)\\
R\sin a = \lambda R \sin b
\end{cases}
With some relatively easy algebra manipulations, we have the solution for the upper intersection as follows.
\begin{cases}
\cos a = \frac{1}{2\lambda}\\
\cos b = \frac{1-2\lambda^2}{2\lambda^2}\\
\sin a=\frac{\sqrt{4\lambda^2 -1}}{2\lambda}\\
\sin b =\frac{\sqrt{4\lambda^2 -1}}{2\lambda^2} 
\end{cases}
Next, find the direction vector of the tangents as follows.
\begin{cases}
\vec{t}_1=\left.\frac{\mathrm{d}\vec{r}_1}{\mathrm{d}\alpha}\right|_{\alpha=a}=R\left(-\sin a, \cos a\right)\\
\vec{t}_2=\left.\frac{\mathrm{d}\vec{r}_2}{\mathrm{d}\beta}\right|_{\beta=b}=\lambda R\left(-\sin b, \cos b\right)
\end{cases}
Now find the angle between the tangents.
\begin{align}
\vec{t}_1 \cdot \vec{t}_2 &= |\vec{t}_1| |\vec{t}_2| \cos \theta \\
\lambda R^2 \left(\cos a \cos b + \sin a \sin b \right)
&=  \lambda R^2  \cos \theta\\
\cos \theta 
&=\cos a \cos b + \sin a \sin b\\
&= \frac{1}{2\lambda} \times \frac{1-2\lambda^2}{2\lambda^2} + \frac{\sqrt{4\lambda^2 -1}}{2\lambda}\times \frac{\sqrt{4\lambda^2 -1}}{2\lambda^2}\\
&= \frac{1-2\lambda^2+4\lambda^2-1}{4\lambda^3}\\
&=\frac{2\lambda^2}{4\lambda^3}\\
&=\frac{1}{2\lambda}
\end{align}
They can be perpendicular each other (or $\theta=90^\circ$) if one of the circles has an infinite radius ($\lambda \to \infty$).

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