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I have been trying to find the angle between two parametric curves:

$$ \begin{array}{lc} C_1 : & \left( \dfrac{a+2}{1+t^2}, \dfrac{a+\sqrt{3}}{1+t^2} \right) \\[5pt] C_2 : & \left( a\cos t, a\sin t \right) \end{array}$$

But I ran into the problem of figuring out the point of intersection. I've tried equating $x(t)$ functions, but I can't solve this kind of equation, when we have a variable inside both trigonometric and polynomial functions. So, should I learn how to solve this equation( hereby push me into right direction, please) or is there another trick about finding the desired angle? Thank you in advance.

Raw pictures

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  • $\begingroup$ Is $C_1$ wrote truly? $\endgroup$ – Nosrati Jan 6 '17 at 19:10
  • $\begingroup$ I feel the denominators in $C_1$ are not equal and possibly an exponent is $3$ instead of $2$. $\endgroup$ – Piquito Jan 6 '17 at 19:11
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Note that $C_2$ is a circle of radius $|a|$ centred at the origin.

Note that $C_1$ is a line segment.

  • $t^2=0 \implies (x,y)=(a+2,a+\sqrt{3})$

  • $t^2 \to \infty \implies (x,y) \to (0,0)$

  • intersection occurs when

\begin{align*} (a+2)^2+(a+\sqrt{3})^2 & \ge a^2 \\ a^2+2(2+\sqrt{3})a+7 & \ge 0 \\ \end{align*}

$$a \le -(2+\sqrt{3}+2\sqrt[4]{3}) \quad \text{or} \quad a \ge -2-\sqrt{3}+2\sqrt[4]{3}$$

Since the line segment $C_1$ approaches to the origin, the angle of intersection is $90^{\circ}$ provided the range of $a$ mentioned above is satisfied.

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HINT: $$(x,y)=\left( \frac{a+2}{1+t^2}, \frac{a+\sqrt{3}}{1+t^2} \right)\Rightarrow \frac{x}{y}=\frac{a+2}{a+\sqrt{3}}.$$ $$(x,y)=\left( a\cos t, a\sin t \right)\Rightarrow x^2+y^2=a^2.$$

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