If l is the length of the intercept made by a common tangent to the circle `x^2+y^2=16` and the ellipse `(x^2)/(25)+(y^2)/(4)=1`, on the coordinate axes, then `81l^2+3` is equal to

Options

A. 5394
B. 5295
C. 5296
D. 5495

Answer

To solve the problem, we need to find the length of the intercept made by a common tangent to the given circle and ellipse on the coordinate axes, and then compute the expression \(81l^2 + 3\).

### Step-by-Step Solution:

1. **Identify the equations of the conics**:
- Circle: \(x^2 + y^2 = 16\)
- Ellipse: \(\frac{x^2}{25} + \frac{y^2}{4} = 1\)

2. **Determine the parameters of the conics**:
- For the circle, the radius \(R\) is \(4\) (since \(R^2 = 16\)).
- For the ellipse, \(a = 5\) and \(b = 2\) (from the standard form of the ellipse).

3. **Write the equation of the tangent line**:
The equation of the tangent to the ellipse can be expressed as:
\[ y = mx \pm \sqrt{a^2m^2 + b^2} \]
Here, we will use the positive root since we are considering the common tangent that extends outward.

4. **Calculate the distance from the center of the circle to the tangent line**:
The distance \(D\) from the center of the circle (0, 0) to the line \(y = mx + \sqrt{25m^2 + 4}\) is given by:
\[ D = \frac{|0 + 0 + \sqrt{25m^2 + 4}|}{\sqrt{1 + m^2}} = \frac{\sqrt{25m^2 + 4}}{\sqrt{1 + m^2}} \]
Set this equal to the radius of the circle:
\[ \frac{\sqrt{25m^2 + 4}}{\sqrt{1 + m^2}} = 4 \]

5. **Square both sides to eliminate the square root**:
\[ 25m^2 + 4 = 16(1 + m^2) \]
Simplifying gives:
\[ 25m^2 + 4 = 16 + 16m^2 \]
\[ 9m^2 = 12 \implies m^2 = \frac{12}{9} = \frac{4}{3} \]
Thus, \(m = \pm \frac{2}{\sqrt{3}}\).

6. **Substituting \(m\) back to find the intercepts**:
Using \(m = \frac{2}{\sqrt{3}}\):
\[ y = \frac{2}{\sqrt{3}}x + \sqrt{25\left(\frac{2}{\sqrt{3}}\right)^2 + 4} \]
Calculate:
\[ = \sqrt{25 \cdot \frac{4}{3} + 4} = \sqrt{\frac{100}{3} + 4} = \sqrt{\frac{100}{3} + \frac{12}{3}} = \sqrt{\frac{112}{3}} = \frac{4\sqrt{7}}{\sqrt{3}} \]

7. **Finding the x-intercept and y-intercept**:
- **x-intercept**: Set \(y = 0\):
\[ 0 = \frac{2}{\sqrt{3}}x + \frac{4\sqrt{7}}{\sqrt{3}} \implies x = -2\sqrt{7} \]
- **y-intercept**: Set \(x = 0\):
\[ y = \frac{4\sqrt{7}}{\sqrt{3}} \]

8. **Length of the intercept \(l\)**:
The length of the intercept \(l\) on the axes is given by:
\[ l = \left| -2\sqrt{7} \right| + \left| \frac{4\sqrt{7}}{\sqrt{3}} \right| = 2\sqrt{7} + \frac{4\sqrt{7}}{\sqrt{3}} \]

9. **Calculate \(l^2\)**:
\[ l^2 = \left(2\sqrt{7} + \frac{4\sqrt{7}}{\sqrt{3}}\right)^2 = 4 \cdot 7 + 2 \cdot 2\sqrt{7} \cdot \frac{4\sqrt{7}}{\sqrt{3}} + \left(\frac{4\sqrt{7}}{\sqrt{3}}\right)^2 \]
This simplifies to:
\[ = 28 + \frac{16 \cdot 7}{3} + \frac{16 \cdot 7}{3} = 28 + \frac{32 \cdot 7}{3} = 28 + \frac{224}{3} \]

10. **Final calculation**:
\[ 81l^2 + 3 = 81\left(28 + \frac{224}{3}\right) + 3 = 81 \cdot 28 + 81 \cdot \frac{224}{3} + 3 \]
Calculate \(81 \cdot 28 = 2268\) and \(81 \cdot \frac{224}{3} = 6048\), so:
\[ 81l^2 + 3 = 2268 + 6048 + 3 = 8319 \]

### Final Answer:
\[ \boxed{8319} \]