The locus of the point of intersection of the lines `x="a"(1+t^(2))/(1-t^(2)), y=(2at)/(1-t^(2))` is a circle of radius a,t being parameter. Is it true or false ?

Answer

To determine whether the locus of the point of intersection of the lines given by the equations

\[ x = a \frac{1 + t^2}{1 - t^2} \]
\[ y = \frac{2at}{1 - t^2} \]

is a circle of radius \( a \), we will follow these steps:

### Step 1: Rewrite the equations
We have the equations for \( x \) and \( y \) in terms of the parameter \( t \).

### Step 2: Square both equations
We will square both equations to eliminate the parameter \( t \).

\[ x = a \frac{1 + t^2}{1 - t^2} \implies x^2 = a^2 \left( \frac{1 + t^2}{1 - t^2} \right)^2 \]

\[ y = \frac{2at}{1 - t^2} \implies y^2 = \left( \frac{2at}{1 - t^2} \right)^2 \]

### Step 3: Combine the equations
Now, we will add the squared equations:

\[ x^2 + y^2 = a^2 \left( \frac{(1 + t^2)^2}{(1 - t^2)^2} \right) + \left( \frac{2at}{1 - t^2} \right)^2 \]

### Step 4: Simplify the right-hand side
We can simplify the right-hand side:

1. The first term becomes:
\[ a^2 \frac{(1 + t^2)^2}{(1 - t^2)^2} \]

2. The second term becomes:
\[ \frac{4a^2t^2}{(1 - t^2)^2} \]

Combining these gives:
\[ x^2 + y^2 = a^2 \frac{(1 + t^2)^2 + 4t^2}{(1 - t^2)^2} \]

### Step 5: Analyze the numerator
The numerator simplifies to:
\[ (1 + t^2)^2 + 4t^2 = 1 + 2t^2 + t^4 + 4t^2 = 1 + 6t^2 + t^4 \]

### Step 6: Final expression
Thus, we have:
\[ x^2 + y^2 = a^2 \frac{1 + 6t^2 + t^4}{(1 - t^2)^2} \]

### Step 7: Determine if it represents a circle
For the locus to represent a circle of radius \( a \), we need:
\[ x^2 + y^2 = a^2 \]

However, the right-hand side involves \( t \) in a non-linear way. The term \( 1 + 6t^2 + t^4 \) cannot be simplified to a constant value. Therefore, the equation does not represent a circle of radius \( a \).

### Conclusion
The statement that the locus of the point of intersection of the lines is a circle of radius \( a \) is **false**.

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