The chords of contact of the pair of tangents drawn from each point on the line 2x+y=4 to the circle `x^2+y^2=1` pass through a fixed point

Options

A. (2,4)
B. (-1/2,-1/4)
C. (1/2,1/4)
D. (-2,-4)

Answer

To solve the problem step by step, we will find the fixed point through which the chords of contact of the pair of tangents drawn from each point on the line \(2x + y = 4\) to the circle \(x^2 + y^2 = 1\) pass.

### Step 1: Identify the Circle and the Line
The equation of the circle is given by:
\[ x^2 + y^2 = 1 \]
The equation of the line is:
\[ 2x + y = 4 \]

### Step 2: Parameterize a Point on the Line
Let’s assume a point \((\alpha, \beta)\) on the line. From the line equation, we can express \(\beta\) in terms of \(\alpha\):
\[ \beta = 4 - 2\alpha \]
Thus, the point can be represented as:
\[ (\alpha, 4 - 2\alpha) \]

### Step 3: Write the Equation of the Chord of Contact
The chord of contact from a point \((x_1, y_1)\) to the circle \(x^2 + y^2 = a^2\) is given by:
\[ xx_1 + yy_1 = a^2 \]
In our case, \(a^2 = 1\), so the equation becomes:
\[ xx_1 + yy_1 = 1 \]
Substituting \(x_1 = \alpha\) and \(y_1 = 4 - 2\alpha\):
\[ x\alpha + y(4 - 2\alpha) = 1 \]

### Step 4: Rearranging the Equation
Expanding the equation gives:
\[ x\alpha + 4y - 2\alpha y = 1 \]
Rearranging this, we have:
\[ \alpha x - 2\alpha y + 4y - 1 = 0 \]
This can be rewritten as:
\[ \alpha(x - 2y) + 4y - 1 = 0 \]

### Step 5: Identify the Fixed Point
For the equation to represent a fixed point, the coefficients of \(\alpha\) must yield a consistent solution regardless of the value of \(\alpha\). This means we can separate the terms involving \(\alpha\) from the constant terms:
1. The coefficient of \(\alpha\) gives us the line \(x - 2y = 0\).
2. The constant term gives us \(4y - 1 = 0\).

From \(4y - 1 = 0\):
\[ y = \frac{1}{4} \]

Substituting \(y = \frac{1}{4}\) into \(x - 2y = 0\):
\[ x - 2\left(\frac{1}{4}\right) = 0 \implies x = \frac{1}{2} \]

### Step 6: Conclusion
Thus, the fixed point through which the chords of contact pass is:
\[ \left(\frac{1}{2}, \frac{1}{4}\right) \]

### Final Answer
The fixed point is \(\left(\frac{1}{2}, \frac{1}{4}\right)\).

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