If the locus of the point of intersection of perpendicular tangents to the ellipse `x^2/a^2+y^2/b^2=1` is a circle with centre at (0,0), then the radius of the circle would be

Options

A. `a^2+b^2`
B. `ab`
C. `b/a`
D. `sqrt(a^2+b^2)`

Answer

To solve the problem, we need to find the radius of the circle that is the locus of the point of intersection of perpendicular tangents to the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given an ellipse and need to find the locus of the intersection points of perpendicular tangents to this ellipse. The locus is said to be a circle centered at the origin (0,0).

2. **Equation of the Ellipse**:
The equation of the ellipse is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

3. **Properties of Perpendicular Tangents**:
The intersection of two perpendicular tangents to the ellipse will lie on a circle known as the director circle of the ellipse. The equation of the director circle for the ellipse is given by:
\[ x^2 + y^2 = a^2 + b^2 \]

4. **Finding the Radius of the Circle**:
The equation of the circle can be rewritten in the standard form:
\[ x^2 + y^2 = r^2 \]
where \( r^2 = a^2 + b^2 \).

Therefore, the radius \( r \) of the circle is:
\[ r = \sqrt{a^2 + b^2} \]

5. **Conclusion**:
The radius of the circle, which is the locus of the point of intersection of the perpendicular tangents to the given ellipse, is:
\[ \sqrt{a^2 + b^2} \]

### Final Answer:
The radius of the circle is \( \sqrt{a^2 + b^2} \).

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