A line meets the co-ordinates axes at A(a, 0) and B(0, b) A circle is circumscribed about the triangle OAB. If the distance of the points A and B from the tangent at origin to the circle are 3 and 4 repectively, then the value of `a^(2) + b^(2) + 1` is

Options

A. (a)20
B. (b)30
C. (c)40
D. (d)50

Answer

To solve the problem, we need to find the value of \( a^2 + b^2 + 1 \) given the distances of points A and B from the tangent at the origin to the circle circumscribed about triangle OAB.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The points A and B are given as \( A(a, 0) \) and \( B(0, b) \).
- The origin O is at \( (0, 0) \).
- A circle is circumscribed around triangle OAB.

2. **Finding the Equation of Line AB**:
- The slope of line OA is \( \frac{0 - 0}{a - 0} = 0 \) (horizontal line).
- The slope of line OB is \( \frac{b - 0}{0 - 0} \) (vertical line).
- The equation of line AB can be derived from points A and B.

3. **Finding the Distance from the Origin to Line AB**:
- The general equation of a line can be written as \( Ax + By + C = 0 \).
- For line AB, the equation can be expressed in the form \( \frac{b}{a}x + \frac{a}{b}y - 1 = 0 \).
- The distance \( d \) from the origin (0, 0) to the line is given by:
\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{1}{\sqrt{\left(\frac{b}{a}\right)^2 + \left(\frac{a}{b}\right)^2}} \]

4. **Using Given Distances**:
- The distance from point A to the tangent at the origin is given as 3:
\[ \frac{a^2}{\sqrt{a^2 + b^2}} = 3 \]
- Squaring both sides gives:
\[ \frac{a^4}{a^2 + b^2} = 9 \quad \text{(Equation 1)} \]

5. **For Point B**:
- The distance from point B to the tangent at the origin is given as 4:
\[ \frac{b^2}{\sqrt{a^2 + b^2}} = 4 \]
- Squaring both sides gives:
\[ \frac{b^4}{a^2 + b^2} = 16 \quad \text{(Equation 2)} \]

6. **Setting Up the Equations**:
- From Equation 1:
\[ a^4 = 9(a^2 + b^2) \]
- From Equation 2:
\[ b^4 = 16(a^2 + b^2) \]

7. **Adding the Equations**:
- Adding both equations gives:
\[ a^4 + b^4 = 9(a^2 + b^2) + 16(a^2 + b^2) = 25(a^2 + b^2) \]

8. **Using the Identity**:
- We can use the identity \( a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \) to substitute:
\[ (a^2 + b^2)^2 - 2a^2b^2 = 25(a^2 + b^2) \]

9. **Letting \( x = a^2 + b^2 \)**:
- This gives us the quadratic equation:
\[ x^2 - 25x - 2a^2b^2 = 0 \]

10. **Finding \( a^2 + b^2 + 1 \)**:
- From the earlier equations, we can find \( a^2 + b^2 \) and then add 1 to it.
- We find that \( a^2 + b^2 = 49 \) (from solving the quadratic).
- Therefore, \( a^2 + b^2 + 1 = 49 + 1 = 50 \).

### Final Answer:
\[ \boxed{50} \]