Let C be the set of curves having the property that the point of intersection of tangent with y-axis is equidistant from the point of tangency and origin (0,0)
If `C_(1),C_(2) in C` such that
`C_(1):` Curve is passing through `(1,0)`
`C_(2):` Curve is passing through `(-1,0)`
The number of common tangents for `C_(1) and C_(2)` is

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the problem, we need to find the number of common tangents between the two curves \( C_1 \) and \( C_2 \) that satisfy the given conditions. Let's go through the solution step by step.

### Step 1: Understanding the curves
The curves \( C_1 \) and \( C_2 \) are defined such that the point of intersection of the tangent with the y-axis is equidistant from the point of tangency and the origin (0,0).

### Step 2: Deriving the equation of the curves
From the problem, we can derive the general equation of the curves. The distance condition gives us the relationship:
\[ OM = MP \]
where \( O \) is the origin, \( M \) is the point where the tangent intersects the y-axis, and \( P \) is the point of tangency.

The equation of the tangent at point \( (x, y) \) is given by:
\[ y - y_0 = m(x - x_0) \]
where \( m \) is the slope of the tangent.

### Step 3: Finding the slope of the tangent
From the distance condition, we can derive the slope \( m \) as follows:
\[ m = \frac{y^2 - x^2}{2xy} \]
This gives us the relationship between \( y \) and \( x \) for the curves.

### Step 4: Substituting \( y = vx \)
Let \( y = vx \). Then, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting this into our slope equation, we get:
\[ v + x\frac{dv}{dx} = \frac{v^2 - 1}{2v} \]

### Step 5: Separating variables and integrating
Rearranging gives us:
\[ 2v \, dv = -(1 + v^2) \frac{dx}{x} \]
Integrating both sides leads to:
\[ \ln(1 + v^2) = -\ln x + C \]
This simplifies to:
\[ 1 + v^2 = \frac{C}{x} \]
Substituting back \( v = \frac{y}{x} \) gives:
\[ 1 + \frac{y^2}{x^2} = \frac{C}{x} \]
Multiplying through by \( x^2 \) leads to:
\[ x^2 + y^2 = Cx \]
This represents a family of circles.

### Step 6: Finding specific curves \( C_1 \) and \( C_2 \)
1. For \( C_1 \) passing through \( (1, 0) \):
\[ 1 + 0 = C \implies C = 1 \implies x^2 + y^2 = x \]
This is a circle centered at \( \left(\frac{1}{2}, 0\right) \) with radius \( \frac{1}{2} \).

2. For \( C_2 \) passing through \( (-1, 0) \):
\[ 1 + 0 = -C \implies C = -1 \implies x^2 + y^2 + x = 0 \]
This is a circle centered at \( \left(-\frac{1}{2}, 0\right) \) with radius \( \frac{1}{2} \).

### Step 7: Analyzing the common tangents
Now we have two circles:
- Circle \( C_1 \): Center \( \left(\frac{1}{2}, 0\right) \), radius \( \frac{1}{2} \)
- Circle \( C_2 \): Center \( \left(-\frac{1}{2}, 0\right) \), radius \( \frac{1}{2} \)

The distance between the centers is:
\[ \text{Distance} = \left| \frac{1}{2} - \left(-\frac{1}{2}\right) \right| = 1 \]
Since the distance between the centers is equal to the sum of the radii (\( \frac{1}{2} + \frac{1}{2} = 1 \)), the circles are externally tangent.

### Conclusion
The number of common tangents between the two curves \( C_1 \) and \( C_2 \) is 3 (two external tangents and one common tangent along the y-axis).