Determine the angle at which the curves x²+y²=2 and y=x²

Determine the angle at which the curves $x^{2}+y^{2}=2$ and $y=x^{2}$ intersect.

Step 1: Intersection points

\[ x^{2}+x^{4}=2 \;\Longrightarrow\; (x^{2}-1)(x^{2}+2)=0 \;\Longrightarrow\; x=\pm1,\ y=1. \tag{1} \]

By symmetry it suffices to work at $P_{1}=(1,1)$.

Step 2: Slopes at $P_1$

Parabola: $y'=2x \Rightarrow k_{1}=2$. Circle (implicit): $2x+2yy'=0 \Rightarrow y'=-x/y \Rightarrow k_{2}=-1$.

Step 3: Angle between the curves

\[ \tan\theta = \left|\frac{k_{1}-k_{2}}{1+k_{1}k_{2}}\right| = \left|\frac{2-(-1)}{1+2(-1)}\right| = \left|\frac{3}{-1}\right| = 3. \tag{2} \]

Step 4: Verification

\[\text{At } P_2=(-1,1)\text{: } k_1=-2,\ k_2=1,\ \tan\theta=\left|\tfrac{-2-1}{1-2}\right|=3\ \checkmark\text{ (same angle by symmetry)}\]