A pair of tangents are drawn from a point P to the circle `x^(2)+y^(2)=1`. If the tangents make an intercept of 2 on the line x=1, the locus of P is

Options

A. straight line
B. pair of lines
C. circle
D. parabola

Answer

To find the locus of the point \( P(h, k) \) from which tangents are drawn to the circle defined by the equation \( x^2 + y^2 = 1 \), given that the tangents make an intercept of 2 on the line \( x = 1 \), we can follow these steps:

### Step 1: Understand the Circle and Tangents
The equation of the circle is \( x^2 + y^2 = 1 \). The center of the circle is at the origin (0, 0) and the radius is 1. A point \( P(h, k) \) lies outside the circle, and we are to find the locus of \( P \).

### Step 2: Use the Tangent Equation
For a point \( P(h, k) \), the equation of the tangents to the circle can be expressed using the formula:
\[ S = T^2 \]
where \( S \) is the equation of the circle, and \( T \) is the equation of the tangents. Here, we have:
\[ S: h^2 + k^2 - 1 = 0 \]
\[ T: xh + yk - (h^2 + k^2 - 1) = 0 \]

### Step 3: Find the Intercept on the Line \( x = 1 \)
Substituting \( x = 1 \) into the tangent equation gives:
\[ 1 \cdot h + y \cdot k - (h^2 + k^2 - 1) = 0 \]
This simplifies to:
\[ h + yk - (h^2 + k^2 - 1) = 0 \]
Rearranging gives:
\[ yk = h^2 + k^2 - 1 - h \]
Thus, the y-intercept can be expressed as:
\[ y = \frac{h^2 + k^2 - 1 - h}{k} \]

### Step 4: Length of the Intercept
The problem states that the tangents make an intercept of 2 on the line \( x = 1 \). This means the distance between the two points where the tangents intersect the line \( x = 1 \) is 2. Therefore, if \( y_1 \) and \( y_2 \) are the intersection points, we have:
\[ y_1 - y_2 = 2 \]

### Step 5: Form a Quadratic Equation
From the tangent equation, we can form a quadratic in \( y \):
\[ k^2y^2 + (2k(h - 1))y + (h^2 - 1) = 0 \]
Using the properties of the roots of the quadratic equation, we know:
1. The sum of the roots \( y_1 + y_2 = -\frac{2k(h - 1)}{k^2} \)
2. The product of the roots \( y_1y_2 = \frac{h^2 - 1}{k^2} \)

### Step 6: Apply the Condition of the Intercept
Using the condition \( y_1 - y_2 = 2 \), we can express this in terms of the sum and product of the roots:
\[ (y_1 - y_2)^2 = (y_1 + y_2)^2 - 4y_1y_2 \]
Substituting the values gives:
\[ 2^2 = \left(-\frac{2k(h - 1)}{k^2}\right)^2 - 4\left(\frac{h^2 - 1}{k^2}\right) \]
This leads to:
\[ 4 = \frac{4k^2(h - 1)^2}{k^4} - \frac{4(h^2 - 1)}{k^2} \]

### Step 7: Simplify and Rearrange
After simplifying, we arrive at the equation:
\[ k^2 = 2h + 2 \]

### Step 8: Substitute for Locus
To express the locus in standard form, we substitute \( h \) and \( k \) with \( x \) and \( y \):
\[ y^2 = 2x + 2 \]
This can be rearranged to:
\[ y^2 - 2x - 2 = 0 \]

### Final Answer
The locus of the point \( P \) is given by:
\[ y^2 = 2x + 2 \]