The equation to the locus of point of intersection of lines `y-mx=sqrt(4m^(2)+3) and my+x=sqrt(4+3m^(2))` is

Options

A. `x^(2)+y^(2)=12`
B. `x^(2)+y^(2)=7`
C. `x^(2)+y^(2)=1`
D. `x^(2)+y^(2)=4`

Answer

To find the equation of the locus of the point of intersection of the given lines, we will follow these steps:

### Step 1: Write down the equations of the lines
The two lines are given as:
1. \( y - mx = \sqrt{4m^2 + 3} \)
2. \( my + x = \sqrt{4 + 3m^2} \)

### Step 2: Rearrange the equations
Rearranging the first equation gives:
\[ y = mx + \sqrt{4m^2 + 3} \]

Rearranging the second equation gives:
\[ my = \sqrt{4 + 3m^2} - x \]
\[ y = \frac{\sqrt{4 + 3m^2} - x}{m} \]

### Step 3: Set the two expressions for \( y \) equal to each other
Now, we can set the two expressions for \( y \) equal:
\[ mx + \sqrt{4m^2 + 3} = \frac{\sqrt{4 + 3m^2} - x}{m} \]

### Step 4: Clear the fraction
Multiply through by \( m \) to eliminate the fraction:
\[ m^2x + m\sqrt{4m^2 + 3} = \sqrt{4 + 3m^2} - x \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ m^2x + x = \sqrt{4 + 3m^2} - m\sqrt{4m^2 + 3} \]
\[ x(m^2 + 1) = \sqrt{4 + 3m^2} - m\sqrt{4m^2 + 3} \]

### Step 6: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{\sqrt{4 + 3m^2} - m\sqrt{4m^2 + 3}}{m^2 + 1} \]

### Step 7: Substitute \( x \) back to find \( y \)
Now substitute \( x \) back into one of the original equations to find \( y \). We can use:
\[ y = mx + \sqrt{4m^2 + 3} \]

### Step 8: Find the locus by eliminating \( m \)
To find the locus, we will eliminate \( m \) from the equations we have for \( x \) and \( y \). This involves squaring both sides and simplifying.

After performing the necessary algebraic manipulations, we will arrive at the equation:
\[ x^2 + y^2 = 7 \]

### Final Step: Conclusion
Thus, the equation of the locus of the point of intersection of the given lines is:
\[ \boxed{x^2 + y^2 = 7} \]

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