Let E be the ellipse `(x^(2))/(9)+(y^(2))/(4)=1` and C be the circle `x^(2)+y^(2)=9`. Let P and Q be the points (1,2) and (2,1) respectively . Then

Options

A. Q lies inside C but outside E
B. Q lies outside both C and E
C. P lies inside both C and E
D. P lies inside C but outside E

Answer

To solve the problem, we will determine whether the points P(1, 2) and Q(2, 1) are inside or outside the given ellipse and circle.

### Step 1: Define the equations
The equations of the ellipse E and the circle C are:
- Ellipse: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\)
- Circle: \(x^2 + y^2 = 9\)

### Step 2: Check point P(1, 2)

#### For the Circle:
1. Substitute the coordinates of point P into the circle's equation:
\[ S_1 = x^2 + y^2 - 9 = 1^2 + 2^2 - 9 = 1 + 4 - 9 = -4 \]
2. Since \(S_1 < 0\), point P is **inside the circle**.

#### For the Ellipse:
1. Substitute the coordinates of point P into the ellipse's equation:
\[ S_2 = \frac{x^2}{9} + \frac{y^2}{4} - 1 = \frac{1^2}{9} + \frac{2^2}{4} - 1 = \frac{1}{9} + \frac{4}{4} - 1 = \frac{1}{9} + 1 - 1 = \frac{1}{9} \]
2. Since \(S_2 > 0\), point P is **outside the ellipse**.

### Step 3: Check point Q(2, 1)

#### For the Circle:
1. Substitute the coordinates of point Q into the circle's equation:
\[ S_3 = x^2 + y^2 - 9 = 2^2 + 1^2 - 9 = 4 + 1 - 9 = -4 \]
2. Since \(S_3 < 0\), point Q is **inside the circle**.

#### For the Ellipse:
1. Substitute the coordinates of point Q into the ellipse's equation:
\[ S_4 = \frac{x^2}{9} + \frac{y^2}{4} - 1 = \frac{2^2}{9} + \frac{1^2}{4} - 1 = \frac{4}{9} + \frac{1}{4} - 1 \]
To combine the fractions, find a common denominator (36):
\[ S_4 = \frac{16}{36} + \frac{9}{36} - \frac{36}{36} = \frac{16 + 9 - 36}{36} = \frac{-11}{36} \]
2. Since \(S_4 < 0\), point Q is **inside the ellipse**.

### Summary of Results:
- Point P(1, 2): Inside Circle, Outside Ellipse
- Point Q(2, 1): Inside Circle, Inside Ellipse

### Final Conclusion:
- P lies inside the circle and outside the ellipse.
- Q lies inside both the circle and the ellipse.