A straight lien is drawn from a fixed point `O` metting a fixed straight line `P`. A point `Q` is taken on the line `OP` such that `OP.OQ` is constant. Show that the locus of `Q` is a circle.

Answer

To show that the locus of the point \( Q \) is a circle given that \( OP \cdot OQ \) is constant, we can follow these steps:

### Step 1: Define the Points and Lines
Let \( O \) be the origin, \( O(0, 0) \). Let \( P \) be a fixed point on a line \( L \) defined by the equation \( y = mx + c \). Assume the coordinates of point \( P \) are \( ( \alpha, \beta ) \).

### Step 2: Define Point \( Q \)
Point \( Q \) lies on the line \( OP \). We can express the coordinates of \( Q \) in terms of a parameter \( t \):
\[ Q(t) = (t \cdot \alpha, t \cdot \beta) \]
where \( t \) is a scalar that varies along the line \( OP \).

### Step 3: Calculate Distances \( OP \) and \( OQ \)
The distance \( OP \) can be calculated as:
\[ OP = \sqrt{\alpha^2 + \beta^2} \]
The distance \( OQ \) is:
\[ OQ = \sqrt{(t \cdot \alpha)^2 + (t \cdot \beta)^2} = t \sqrt{\alpha^2 + \beta^2} \]

### Step 4: Set Up the Given Condition
According to the problem, \( OP \cdot OQ \) is constant. Let this constant be \( k \):
\[ OP \cdot OQ = \sqrt{\alpha^2 + \beta^2} \cdot (t \sqrt{\alpha^2 + \beta^2}) = k \]
This simplifies to:
\[ (t(\alpha^2 + \beta^2)) = k \]
From this, we can express \( t \):
\[ t = \frac{k}{\alpha^2 + \beta^2} \]

### Step 5: Substitute \( t \) in Coordinates of \( Q \)
Substituting \( t \) back into the coordinates of \( Q \):
\[ Q = \left( \frac{k \cdot \alpha}{\alpha^2 + \beta^2}, \frac{k \cdot \beta}{\alpha^2 + \beta^2} \right) \]

### Step 6: Express the Locus of \( Q \)
Let \( x = Q_x = \frac{k \cdot \alpha}{\alpha^2 + \beta^2} \) and \( y = Q_y = \frac{k \cdot \beta}{\alpha^2 + \beta^2} \). We can express \( x \) and \( y \) in terms of \( k \):
\[ x^2 + y^2 = \left(\frac{k \cdot \alpha}{\alpha^2 + \beta^2}\right)^2 + \left(\frac{k \cdot \beta}{\alpha^2 + \beta^2}\right)^2 \]
This simplifies to:
\[ x^2 + y^2 = \frac{k^2 (\alpha^2 + \beta^2)}{(\alpha^2 + \beta^2)^2} = \frac{k^2}{\alpha^2 + \beta^2} \]

### Step 7: Rearranging to Circle Equation
This can be rearranged to form the equation of a circle:
\[ x^2 + y^2 = r^2 \]
where \( r^2 = \frac{k^2}{\alpha^2 + \beta^2} \) is a constant since \( k \) and \( \alpha^2 + \beta^2 \) are constants.

### Conclusion
Thus, the locus of point \( Q \) is indeed a circle.

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