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

Options

A. a) `x^(2)+y^(2)=a^(2)+b^(2)`
B. b) `x^(2)+y^(2)=a^(2)-b^(2)`
C. c) `x^(2)+y^(2)=(a+b)^(2)`
D. d) `x^(2)+y^(2)=(a-b)^(2)`

Answer

To solve the problem of finding the locus of the point of intersection of the perpendicular tangents to the circles defined 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
For the circle \(x^2 + y^2 = a^2\), the equation of the tangent at an angle \(\alpha\) is given by:
\[ x \cos \alpha + y \sin \alpha = a \]
For the circle \(x^2 + y^2 = b^2\), the equation of the tangent that is perpendicular to the first tangent is:
\[ x \sin \alpha - y \cos \alpha = b \]

### Step 2: Set up the equations
Let the point of intersection of these tangents be \((h, k)\). Then, substituting \((h, k)\) into the tangent equations gives us:
1. \(h \cos \alpha + k \sin \alpha = a\) (Equation 1)
2. \(h \sin \alpha - k \cos \alpha = b\) (Equation 2)

### Step 3: Square and add the equations
Now, we will square both equations and add them together:
- Squaring Equation 1:
\[ (h \cos \alpha + k \sin \alpha)^2 = a^2 \]
Expanding this:
\[ h^2 \cos^2 \alpha + 2hk \cos \alpha \sin \alpha + k^2 \sin^2 \alpha = a^2 \]

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

### Step 4: Add the two squared equations
Now, adding the two expanded equations:
\[ (h^2 \cos^2 \alpha + k^2 \sin^2 \alpha + 2hk \cos \alpha \sin \alpha) + (h^2 \sin^2 \alpha + k^2 \cos^2 \alpha - 2hk \sin \alpha \cos \alpha) = a^2 + b^2 \]
This simplifies to:
\[ h^2 (\cos^2 \alpha + \sin^2 \alpha) + k^2 (\sin^2 \alpha + \cos^2 \alpha) = a^2 + b^2 \]

### Step 5: Use the Pythagorean identity
Using the identity \(\cos^2 \alpha + \sin^2 \alpha = 1\), we can simplify the equation:
\[ h^2 + k^2 = a^2 + b^2 \]

### Step 6: Write the locus equation
The above equation represents the locus of the point \((h, k)\). Therefore, replacing \(h\) with \(x\) and \(k\) with \(y\), we get:
\[ x^2 + y^2 = a^2 + b^2 \]

### Final Result
Thus, the locus of the point of intersection of the perpendicular tangents to the circles is:
\[ \boxed{x^2 + y^2 = a^2 + b^2} \]