Locus of the point of intersection of perpendicular tangents to the circle `x^(2)+y^(2)=16` is

Options

A. `x^(2)+y^(2)=8`
B. `x^(2)+y^(2)=32`
C. `x^(2)+y^(2)=64`
D. `x^(2)+y^(2)=16`

Answer

To find the locus of the point of intersection of perpendicular tangents to the circle given by the equation \( x^2 + y^2 = 16 \), we can follow these steps:

### Step 1: Identify the Circle's Properties
The given equation of the circle is \( x^2 + y^2 = 16 \). This can be rewritten in standard form as:
\[ x^2 + y^2 = r^2 \]
where \( r^2 = 16 \), hence the radius \( r = 4 \).

**Hint:** The radius of a circle can be found by taking the square root of the constant term in the equation of the circle.

### Step 2: Understand the Concept of Perpendicular Tangents
For a circle, if two tangents are drawn from an external point and they are perpendicular to each other, the point of intersection of these tangents will lie on a specific locus. This locus is known as the director circle.

**Hint:** The director circle is associated with the concept of tangents and their intersection properties.

### Step 3: Determine the Radius of the Director Circle
The radius of the director circle \( r' \) is given by the formula:
\[ r' = r \sqrt{2} \]
Substituting the radius \( r = 4 \):
\[ r' = 4 \sqrt{2} \]

**Hint:** The radius of the director circle is derived from the original circle's radius and the properties of perpendicular tangents.

### Step 4: Write the Equation of the Director Circle
The equation of the director circle can be written as:
\[ x^2 + y^2 = (r')^2 \]
Substituting \( r' = 4\sqrt{2} \):
\[ x^2 + y^2 = (4\sqrt{2})^2 = 32 \]

**Hint:** The equation of a circle is always in the form \( x^2 + y^2 = r^2 \).

### Conclusion
Thus, the locus of the point of intersection of the perpendicular tangents to the circle \( x^2 + y^2 = 16 \) is given by:
\[ \boxed{x^2 + y^2 = 32} \]