Equation of the tangent to the circle at the point (1, -1) whose centre is the point of intersection of the straight lines x-y=1 and 2x+y-3=0, is

Options

A. `x+ 4y +3 =0`
B. `3x -y -4=0`
C. `x -3y- 4=0`
D. ` 4x +y -3=0`

Answer

To find the equation of the tangent to the circle at the point (1, -1), where the center of the circle is the point of intersection of the lines \(x - y = 1\) and \(2x + y - 3 = 0\), we can follow these steps:

### Step 1: Find the point of intersection of the lines
We need to solve the system of equations given by the two lines:

1. \(x - y = 1\) (Equation 1)
2. \(2x + y - 3 = 0\) (Equation 2)

From Equation 1, we can express \(y\) in terms of \(x\):
\[ y = x - 1 \]

Now, substitute this expression for \(y\) into Equation 2:
\[ 2x + (x - 1) - 3 = 0 \]
\[ 2x + x - 1 - 3 = 0 \]
\[ 3x - 4 = 0 \]
\[ 3x = 4 \quad \Rightarrow \quad x = \frac{4}{3} \]

Now substitute \(x = \frac{4}{3}\) back into the expression for \(y\):
\[ y = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]

Thus, the point of intersection (the center of the circle) is:
\[ \left(\frac{4}{3}, \frac{1}{3}\right) \]

### Step 2: Find the slope of the radius at the point (1, -1)
The slope of the line connecting the center \(\left(\frac{4}{3}, \frac{1}{3}\right)\) to the point (1, -1) can be calculated as follows:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - \frac{1}{3}}{1 - \frac{4}{3}} = \frac{-\frac{3}{3} - \frac{1}{3}}{1 - \frac{4}{3}} = \frac{-\frac{4}{3}}{-\frac{1}{3}} = 4 \]

### Step 3: Find the slope of the tangent
The slope of the tangent line is the negative reciprocal of the slope of the radius:
\[ \text{slope of tangent} = -\frac{1}{4} \]

### Step 4: Use the point-slope form to find the equation of the tangent
Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point (1, -1):
\[ y - (-1) = -\frac{1}{4}(x - 1) \]
\[ y + 1 = -\frac{1}{4}x + \frac{1}{4} \]
\[ y = -\frac{1}{4}x + \frac{1}{4} - 1 \]
\[ y = -\frac{1}{4}x - \frac{3}{4} \]

### Step 5: Rearranging the equation
To express this in standard form:
\[ \frac{1}{4}x + y + \frac{3}{4} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ x + 4y + 3 = 0 \]

Thus, the equation of the tangent to the circle at the point (1, -1) is:
\[ \boxed{x + 4y + 3 = 0} \]