The tangents from P to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` are mutually perpendicular show that the locus of P is the circle `x^(2)+y^(2)=a^(2)-b^(2)`

Answer

To solve the problem, we need to show that the locus of point P, from which the tangents to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are mutually perpendicular, is the circle given by \(x^2 + y^2 = a^2 - b^2\).

### Step-by-Step Solution:

1. **Equation of the Hyperbola**:
The hyperbola is given by the equation:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

2. **Coordinates of Point P**:
Let the coordinates of point P be \(P(h, k)\).

3. **Equation of Tangent to the Hyperbola**:
The equation of the tangent to the hyperbola from point \(P(h, k)\) can be expressed in slope form as:
\[ y - mx = \pm \sqrt{a^2m^2 - b^2} \]
Rearranging gives:
\[ y - mx = \sqrt{a^2m^2 - b^2} \quad \text{and} \quad y - mx = -\sqrt{a^2m^2 - b^2} \]

4. **Substituting Point P into the Tangent Equation**:
Since the tangents pass through point \(P(h, k)\), we substitute \(x = h\) and \(y = k\):
\[ k - mh = \pm \sqrt{a^2m^2 - b^2} \]

5. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ (k - mh)^2 = a^2m^2 - b^2 \]
Expanding the left side:
\[ k^2 - 2kmh + m^2h^2 = a^2m^2 - b^2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ (h^2 - a^2)m^2 - 2kmh + (k^2 + b^2) = 0 \]

7. **Condition for Mutually Perpendicular Tangents**:
For the tangents to be mutually perpendicular, the product of the slopes \(m_1\) and \(m_2\) must equal \(-1\). The slopes are the roots of the quadratic equation:
\[ Am^2 + Bm + C = 0 \]
where \(A = h^2 - a^2\), \(B = -2kh\), and \(C = k^2 + b^2\).

8. **Using the Product of Roots**:
The product of the roots \(m_1 m_2\) is given by:
\[ m_1 m_2 = \frac{C}{A} = -1 \]
Therefore:
\[ \frac{k^2 + b^2}{h^2 - a^2} = -1 \]

9. **Cross-Multiplying**:
Cross-multiplying gives:
\[ k^2 + b^2 = - (h^2 - a^2) \]
Simplifying this leads to:
\[ k^2 + b^2 + h^2 = a^2 \]

10. **Final Form**:
Replacing \(h\) and \(k\) with \(x\) and \(y\) respectively, we have:
\[ x^2 + y^2 = a^2 - b^2 \]

### Conclusion:
Thus, we have shown that the locus of point \(P\) is the circle given by:
\[ x^2 + y^2 = a^2 - b^2 \]