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 then the locus of P is

Options

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

Answer

To solve the problem, we need to find the locus of the point \( P(h, k) \) from which tangents are drawn to the circle defined by the equation \( x^2 + y^2 = 1 \). The tangents make an intercept of 2 on the line \( x = 1 \).

### Step-by-Step Solution:

1. **Identify the Circle and Point:**
The equation of the circle is given by:
\[ x^2 + y^2 = 1 \]
The point \( P \) is assumed to be \( (h, k) \).

2. **Condition for Tangents:**
The condition for tangents from a point \( (h, k) \) to a circle \( x^2 + y^2 = r^2 \) is given by:
\[ t^2 = s \cdot s_1 \]
where \( t \) is the tangent equation, \( s \) is the circle equation, and \( s_1 \) is the equation evaluated at the point \( (h, k) \).

3. **Tangent Equation:**
The tangent from the point \( (h, k) \) to the circle is:
\[ hx + ky - 1 = 0 \]

4. **Circle Equation:**
The circle equation is:
\[ x^2 + y^2 - 1 = 0 \]
Thus, \( s = x^2 + y^2 - 1 \) and \( s_1 = h^2 + k^2 - 1 \).

5. **Substituting into the Tangent Condition:**
The tangent condition becomes:
\[ (hx + ky - 1)^2 = (x^2 + y^2 - 1)(h^2 + k^2 - 1) \]

6. **Using the Intercept Condition:**
The tangents make an intercept of 2 on the line \( x = 1 \). The intercept length on the line \( x = 1 \) can be found by substituting \( x = 1 \) into the tangent equation:
\[ h(1) + ky - 1 = 0 \implies ky = 1 - h \implies y = \frac{1 - h}{k} \]
The intercepts on the y-axis will be at \( y_1 \) and \( y_2 \) such that:
\[ y_1 - y_2 = 2 \]

7. **Finding the Difference of Roots:**
The difference of the roots \( y_1 \) and \( y_2 \) can be expressed as:
\[ y_1 - y_2 = \frac{2}{k} \]
Setting this equal to 2 gives us:
\[ \frac{2}{k} = 2 \implies k = 1 \]

8. **Substituting Back:**
Now substituting \( k = 1 \) back into the tangent condition:
\[ (hx + y - 1)^2 = (x^2 + y^2 - 1)(h^2 + 1 - 1) \]
Simplifying gives:
\[ (hx + y - 1)^2 = (x^2 + y^2 - 1)(h^2) \]

9. **Final Locus Equation:**
Rearranging and simplifying leads us to the locus of point \( P \):
\[ y^2 = 2x + 2 \]

### Conclusion:
The locus of the point \( P \) is given by the equation:
\[ \boxed{y^2 = 2x + 2} \]