The chord of contact of (3,-1) w.r.t the circle `x^(2)+y^(2)+2x-4y+1=0` is

Options

A. `4x-3y+6=0`
B. `4x-3y+1=0`
C. `5x+y+5=0`
D. `4x-3y+11=0`

Answer

To find the chord of contact of the point (3, -1) with respect to the circle given by the equation \(x^2 + y^2 + 2x - 4y + 1 = 0\), we can follow these steps:

### Step 1: Write the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 + 2x - 4y + 1 = 0 \]
We can rewrite it in standard form by completing the square.

1. For \(x\):
\[ x^2 + 2x = (x + 1)^2 - 1 \]
2. For \(y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Now substituting back into the equation:
\[ (x + 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0 \]
This simplifies to:
\[ (x + 1)^2 + (y - 2)^2 - 4 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x + 1)^2 + (y - 2)^2 = 4 \]
This indicates that the center of the circle is at \((-1, 2)\) and the radius is \(2\).

### Step 2: Use the formula for the chord of contact
The chord of contact for a point \((x_1, y_1)\) with respect to the circle is given by the equation:
\[ xx_1 + yy_1 + 2h = 0 \]
where \(h\) is derived from the circle's equation in the form \(x^2 + y^2 + 2gx + 2fy + c = 0\). Here, \(g = 1\), \(f = -2\), and \(c = 1\).

The equation becomes:
\[ xx_1 + yy_1 + (g x + f y + c) = 0 \]
Substituting \(x_1 = 3\) and \(y_1 = -1\):
\[ x(3) + y(-1) + (1 \cdot 3 - 2 \cdot (-1) + 1) = 0 \]
This simplifies to:
\[ 3x - y + (3 + 2 + 1) = 0 \]
\[ 3x - y + 6 = 0 \]

### Step 3: Final equation of the chord of contact
Thus, the chord of contact of the point \((3, -1)\) with respect to the circle is:
\[ 3x - y + 6 = 0 \]