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But does it intersect in more points? No — since the circle lies entirely on or inside the ellipse in \( y \), but at \( x=0 \), both are 1.
Actually, the circle is inscribed: at \( x=0 \), both have \( y = \pm8 \), and for \( |x| > 0 \), ellipse allows larger \( x \), but circle restricts \( x^2 + y^2 = 64 \), so maximum \( y \) is bounded.
Try solving numerically:
At \( x = 8 \), ellipse: \( y^2 = 64(1 - 64/100) = 64(0.36) = 23.04 \), so \( y \approx 4.8 \).
Circle: at \( x=8 \), \( y=0 \), so not intersecting.
Try solving:
From earlier:
x^2\left( \frac{1}{100} - \frac{1}{64} \right) = 0 \Rightarrow x^2 = 0
Only solution is \( x = 0 \), \( y = \pm8 \).
But geometrically: ellipse \( \frac{x^2}{100} + \frac{y^2}{64} = 1 \) has semi-major axis 10 (x), semi-minor 8 (y).