The locus of point of intersection of the lines `y+mx=sqrt(a^2m^2+b^2)` and `my-x=sqrt(a^2+b^2m^2)` is

Options

A. `x^2+y^2=(1)/(a^2)+(1)/(b^2)`
B. `x^2+y^2=a^2+b^2`
C. `x^2+y^2=a^2-b^2`
D. `1/x^2+1/y^2=a^2-b^2`

Explanation

To find the locus of the point of intersection of the lines given by the equations \(y + mx = \sqrt{a^2 m^2 + b^2}\) and \(my - x = \sqrt{a^2 + b^2 m^2}\), we will follow these steps:

### Step 1: Rewrite the equations
The first equation can be rearranged to express \(y\):
\[ y = \sqrt{a^2 m^2 + b^2} - mx \]
The second equation can be rearranged to express \(x\):
\[ x = my - \sqrt{a^2 + b^2 m^2} \]

### Step 2: Substitute \(y\) from the first equation into the second
Substituting \(y\) from the first equation into the second gives:
\[ x = m\left(\sqrt{a^2 m^2 + b^2} - mx\right) - \sqrt{a^2 + b^2 m^2} \]
Expanding this, we have:
\[ x = m\sqrt{a^2 m^2 + b^2} - m^2x - \sqrt{a^2 + b^2 m^2} \]

### Step 3: Rearranging the equation
Rearranging the equation to isolate \(x\):
\[ x + m^2x = m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2} \]
Factoring out \(x\):
\[ x(1 + m^2) = m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2} \]
Thus,
\[ x = \frac{m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2}}{1 + m^2} \]

### Step 4: Substitute \(x\) back to find \(y\)
Now substitute this expression for \(x\) back into the equation for \(y\):
\[ y = \sqrt{a^2 m^2 + b^2} - m\left(\frac{m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2}}{1 + m^2}\right) \]

### Step 5: Simplifying the expression for \(y\)
After substituting and simplifying, we will have an expression for \(y\) in terms of \(m\):
\[ y = \frac{\sqrt{a^2 m^2 + b^2}(1 + m^2) - m(m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2})}{1 + m^2} \]

### Step 6: Square both expressions
To find the locus, we will square both \(x\) and \(y\) and add them:
\[ x^2 + y^2 = \left(\frac{m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2}}{1 + m^2}\right)^2 + \left(\frac{\sqrt{a^2 m^2 + b^2}(1 + m^2) - m(m\sqrt{a^2 m^2 + b^2} - \sqrt{a^2 + b^2 m^2})}{1 + m^2}\right)^2 \]

### Step 7: Final expression
After simplifying the above expression, we will arrive at:
\[ x^2 + y^2 = a^2 + b^2 \]

### Conclusion
Thus, the locus of the point of intersection of the given lines is:
\[ \boxed{x^2 + y^2 = a^2 + b^2} \]