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 \( 81l^2 + 3 \).

### 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 ellipse**:
- The semi-major axis \( a = 5 \) (since \( a^2 = 25 \))
- The semi-minor axis \( b = 2 \) (since \( b^2 = 4 \))

3. **Equation of the tangent to the ellipse**:
The equation of the tangent to the ellipse at slope \( m \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
Substituting \( a = 5 \) and \( b = 2 \):
\[ y = mx \pm \sqrt{25m^2 + 4} \]

4. **Distance from the origin to the tangent line**:
The distance \( d \) from the origin (0,0) to the line \( y = mx + c \) is given by:
\[ d = \frac{|c|}{\sqrt{1 + m^2}} \]
For the tangent to be a common tangent to the circle, this distance must equal the radius of the circle, which is 4:
\[ 4 = \frac{\sqrt{25m^2 + 4}}{\sqrt{1 + m^2}} \]

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

6. **Find the slopes**:
\[ m = \pm \frac{2}{\sqrt{3}} \]

7. **Substituting back to find the equation of the tangent**:
Using \( m = \frac{2}{\sqrt{3}} \):
\[ y = \frac{2}{\sqrt{3}}x \pm \sqrt{25\left(\frac{4}{3}\right) + 4} \]
Calculating the square root term:
\[ = \sqrt{\frac{100}{3} + 4} = \sqrt{\frac{100}{3} + \frac{12}{3}} = \sqrt{\frac{112}{3}} = \frac{4\sqrt{7}}{\sqrt{3}} \]
So the equation of the tangent is:
\[ y = \frac{2}{\sqrt{3}}x \pm \frac{4\sqrt{7}}{\sqrt{3}} \]

8. **Finding intercepts on the axes**:
- For \( y = \frac{2}{\sqrt{3}}x + \frac{4\sqrt{7}}{\sqrt{3}} \):
- Set \( y = 0 \) to find the x-intercept:
\[ 0 = \frac{2}{\sqrt{3}}x + \frac{4\sqrt{7}}{\sqrt{3}} \implies x = -2\sqrt{7} \]
- Set \( x = 0 \) to find the y-intercept:
\[ y = \frac{4\sqrt{7}}{\sqrt{3}} \]

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

10. **Calculate \( l^2 \)**:
\[ l^2 = \left(2\sqrt{7} + \frac{4\sqrt{7}}{\sqrt{3}}\right)^2 \]
Simplifying this expression will yield \( l^2 \).

11. **Final Calculation**:
Compute \( 81l^2 + 3 \).