The locus of th point (l,m). If the line lx+my=1 touches the circle `x^(2)+y^(2)=a^(2)` is

Options

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

Answer

To find the locus of the point (l, m) such that the line \( lx + my = 1 \) touches the circle \( x^2 + y^2 = a^2 \), we can follow these steps:

### Step 1: Understanding the condition for tangency
For the line \( lx + my = 1 \) to touch the circle \( x^2 + y^2 = a^2 \), the perpendicular distance from the center of the circle (which is at the origin (0, 0)) to the line must equal the radius of the circle \( a \).

### Step 2: Finding the perpendicular distance from the origin to the line
The formula for the perpendicular distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( lx + my = 1 \), we can rewrite it as \( lx + my - 1 = 0 \). Here, \( A = l \), \( B = m \), and \( C = -1 \). The distance from the origin (0, 0) to this line is:
\[ d = \frac{|l(0) + m(0) - 1|}{\sqrt{l^2 + m^2}} = \frac{1}{\sqrt{l^2 + m^2}} \]

### Step 3: Setting the distance equal to the radius
Since the line touches the circle, we set the distance equal to the radius \( a \):
\[ \frac{1}{\sqrt{l^2 + m^2}} = a \]

### Step 4: Squaring both sides
Squaring both sides of the equation gives:
\[ \frac{1}{l^2 + m^2} = a^2 \]
This can be rearranged to:
\[ l^2 + m^2 = \frac{1}{a^2} \]

### Step 5: Rearranging the equation
Multiplying through by \( a^2 \) leads us to:
\[ a^2(l^2 + m^2) = 1 \]
This represents the equation of a circle with radius \( \frac{1}{a} \).

### Conclusion
Thus, the locus of the point \( (l, m) \) is given by:
\[ a^2(x^2 + y^2) = 1 \]

### Final Answer
The correct option is:
\[ a^2(x^2 + y^2) = 1 \]