The areas of two circles are in a ratio of 4:9. If both radii are integers and `r_(1)-r_(2)=2`, which of the following is the radius of the larger circle?

Options

A. `4`
B. `6`
C. `8`
D. `9`

Answer

To solve the problem, we need to find the radius of the larger circle given that the areas of two circles are in the ratio of 4:9, the difference between their radii is 2, and both radii are integers.

### Step-by-step Solution:

1. **Understand the Ratio of Areas**:
The areas of two circles are given in the ratio \(4:9\). If \(r_1\) is the radius of the larger circle and \(r_2\) is the radius of the smaller circle, then:
\[ \frac{\pi r_1^2}{\pi r_2^2} = \frac{4}{9} \]
This simplifies to:
\[ \frac{r_1^2}{r_2^2} = \frac{4}{9} \]

2. **Take the Square Root**:
Taking the square root of both sides gives us:
\[ \frac{r_1}{r_2} = \frac{2}{3} \]
This means that \(r_1 = \frac{2}{3} r_2\).

3. **Use the Difference of Radii**:
We are also given that the difference between the two radii is 2:
\[ r_1 - r_2 = 2 \]

4. **Express \(r_1\) in terms of \(r_2\)**:
From the ratio, we can express \(r_1\) in terms of \(r_2\):
\[ r_1 = \frac{2}{3} r_2 \]
Substitute this into the difference equation:
\[ \frac{2}{3} r_2 - r_2 = 2 \]

5. **Simplify the Equation**:
This simplifies to:
\[ \frac{2}{3} r_2 - \frac{3}{3} r_2 = 2 \]
\[ -\frac{1}{3} r_2 = 2 \]

6. **Solve for \(r_2\)**:
Multiplying both sides by -3 gives:
\[ r_2 = -6 \]
Since \(r_2\) must be a positive integer, we need to reconsider our expressions.

7. **Rearranging the Difference**:
We can rearrange the difference equation:
\[ r_1 = r_2 + 2 \]

8. **Substituting into the Ratio**:
Now substitute \(r_1\) into the ratio equation:
\[ \frac{(r_2 + 2)^2}{r_2^2} = \frac{4}{9} \]
Cross-multiplying gives:
\[ 9(r_2 + 2)^2 = 4r_2^2 \]

9. **Expanding and Rearranging**:
Expanding the left side:
\[ 9(r_2^2 + 4r_2 + 4) = 4r_2^2 \]
\[ 9r_2^2 + 36r_2 + 36 = 4r_2^2 \]
Rearranging gives:
\[ 5r_2^2 + 36r_2 + 36 = 0 \]

10. **Using the Quadratic Formula**:
Now we can use the quadratic formula to solve for \(r_2\):
\[ r_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 5\), \(b = 36\), and \(c = 36\):
\[ r_2 = \frac{-36 \pm \sqrt{36^2 - 4 \cdot 5 \cdot 36}}{2 \cdot 5} \]
\[ = \frac{-36 \pm \sqrt{1296 - 720}}{10} \]
\[ = \frac{-36 \pm \sqrt{576}}{10} \]
\[ = \frac{-36 \pm 24}{10} \]
This gives two possible values:
\[ r_2 = \frac{-12}{10} \quad \text{(not valid)} \quad \text{or} \quad r_2 = \frac{-60}{10} = -6 \quad \text{(not valid)} \]

11. **Finding Integer Values**:
Since we need integer values, we can check integer pairs that satisfy \(r_1 - r_2 = 2\) and the area ratio. Testing integers:
- If \(r_2 = 4\), then \(r_1 = 6\).
- Check the area ratio:
\[ \frac{6^2}{4^2} = \frac{36}{16} = \frac{9}{4} \quad \text{(valid)} \]

Thus, the radius of the larger circle \(r_1\) is **6**.