The locus of the point of intersection of perpendicular tangents to the circles `x^(2)+y^(2)=a^(2)` and `x^(2)+y^(2)=b^(2)` , is

Options

A. `x^(2)+y^(2)=a^(2)-b^(2)`
B. `x^(2)+y^(2)=a^(2)+b^(2)`
C. `x^(2)+y^(2)=(a+b)^(2)`
D. none of these

Explanation

To find the locus of the point of intersection of perpendicular tangents to the circles given by the equations \(x^2 + y^2 = a^2\) and \(x^2 + y^2 = b^2\), we can follow these steps:

### Step 1: Write the equations of the tangents
The equation of the tangent to the circle \(x^2 + y^2 = a^2\) can be expressed as:
\[ x \cos \theta + y \sin \theta = a \]
where \(\theta\) is the angle that the tangent makes with the x-axis.

### Step 2: Write the equation of the perpendicular tangent
Since the tangents are perpendicular, the equation of the tangent to the circle \(x^2 + y^2 = b^2\) can be written as:
\[ x \sin \theta - y \cos \theta = b \]

### Step 3: Set up the equations for the point of intersection
Let the point of intersection of these tangents be \((h, k)\). We can substitute \(x\) with \(h\) and \(y\) with \(k\) in both tangent equations:
1. From the first tangent:
\[ h \cos \theta + k \sin \theta = a \quad \text{(Equation 1)} \]
2. From the second tangent:
\[ h \sin \theta - k \cos \theta = b \quad \text{(Equation 2)} \]

### Step 4: Square both equations and add them
Now, we will square both equations and add them:
1. Squaring Equation 1:
\[ (h \cos \theta + k \sin \theta)^2 = a^2 \]
Expanding this gives:
\[ h^2 \cos^2 \theta + 2hk \cos \theta \sin \theta + k^2 \sin^2 \theta = a^2 \]

2. Squaring Equation 2:
\[ (h \sin \theta - k \cos \theta)^2 = b^2 \]
Expanding this gives:
\[ h^2 \sin^2 \theta - 2hk \sin \theta \cos \theta + k^2 \cos^2 \theta = b^2 \]

### Step 5: Add the two squared equations
Now, adding the two expanded equations:
\[ (h^2 \cos^2 \theta + k^2 \sin^2 \theta + h^2 \sin^2 \theta + k^2 \cos^2 \theta) + (2hk \cos \theta \sin \theta - 2hk \sin \theta \cos \theta) = a^2 + b^2 \]
This simplifies to:
\[ h^2 (\cos^2 \theta + \sin^2 \theta) + k^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we have:
\[ h^2 + k^2 = a^2 + b^2 \]

### Step 6: Write the locus equation
The equation \(h^2 + k^2 = a^2 + b^2\) represents a circle centered at the origin with radius \(\sqrt{a^2 + b^2}\). Therefore, the locus of the point of intersection of the perpendicular tangents is:
\[ x^2 + y^2 = a^2 + b^2 \]

### Final Answer
The locus of the point of intersection of the perpendicular tangents to the circles \(x^2 + y^2 = a^2\) and \(x^2 + y^2 = b^2\) is given by:
\[ x^2 + y^2 = a^2 + b^2 \]