Given: - Ellipse: \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) - Circle: \( x^2 + y^2 = 9 \) - Points \( P(1, 2) \) and \( Q(2, 1) \) ### Step-by-Step Verification: #### 1. **Checking Point \( P(1, 2) \) with respect to the Circle:** The equation of the circle is \( x^2 + y^2 = 9 \). Substitute \( P(1, 2) \): \[ 1^2 + 2^2 = 1 + 4 = 5 < 9 \] Thus, point \( P \) lies inside the circle \( C \). #### 2. **Checking Point \( Q(2, 1) \) with respect to the Circle:** Substitute \( Q(2, 1) \): \[ 2^2 + 1^2 = 4 + 1 = 5 < 9 \] Thus, point \( Q \) also lies inside the circle \( C \). #### 3. **Checking Point \( P(1, 2) \) with respect to the Ellipse:** The equation of the ellipse is \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \). Substitute \( P(1, 2) \): \[ \frac{1^2}{9} + \frac{2^2}{4} = \frac{1}{9} + 1 = \frac{10}{9} > 1 \] Thus, point \( P \) lies **outside** the ellipse \( E \). #### 4. **Checking Point \( Q(2, 1) \) with respect to the Ellipse:** Substitute \( Q(2, 1) \): \[ \frac{2^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{16}{36} + \frac{9}{36} = \frac{25}{36} < 1 \] Thus, point \( Q \) lies **inside** the ellipse \( E \). ### Conclusion: - **Point \( P \)** lies inside the circle \( C \) but outside the ellipse \( E \). - **Point \( Q \)** lies inside both the circle \( C \) and the ellipse \( E \). Therefore, the correct option is: \[ \boxed{D. \text{P lies inside C but outside E.}} \]