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

Explanation

To find the equation of the tangent to the circle at the point (1, -1), we first need to determine the center of the circle, which is located at the intersection of the two given lines.

### Step 1: Find the intersection of the lines
The equations of the lines are:
1. \( x - y = 1 \) (Equation 1)
2. \( 2x + y - 3 = 0 \) (Equation 2)

We can solve these equations simultaneously.

From Equation 1:
\[ y = x - 1 \]

Substituting \( y \) in Equation 2:
\[ 2x + (x - 1) - 3 = 0 \]
\[ 2x + x - 1 - 3 = 0 \]
\[ 3x - 4 = 0 \]
\[ x = \frac{4}{3} \]

Now substituting \( x \) back into Equation 1 to find \( y \):
\[ y = \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]

Thus, the center of the circle is at the point \( \left( \frac{4}{3}, \frac{1}{3} \right) \).

### Step 2: Calculate the radius of the circle
The radius \( r \) is the distance from the center \( \left( \frac{4}{3}, \frac{1}{3} \right) \) to the point \( (1, -1) \).

Using the distance formula:
\[ r = \sqrt{ \left( 1 - \frac{4}{3} \right)^2 + \left( -1 - \frac{1}{3} \right)^2 } \]
Calculating the x-component:
\[ 1 - \frac{4}{3} = \frac{3}{3} - \frac{4}{3} = -\frac{1}{3} \]
Calculating the y-component:
\[ -1 - \frac{1}{3} = -\frac{3}{3} - \frac{1}{3} = -\frac{4}{3} \]

Now substituting these values into the distance formula:
\[ r = \sqrt{ \left( -\frac{1}{3} \right)^2 + \left( -\frac{4}{3} \right)^2 } \]
\[ = \sqrt{ \frac{1}{9} + \frac{16}{9} } = \sqrt{ \frac{17}{9} } = \frac{\sqrt{17}}{3} \]

### Step 3: Write the equation of the tangent
The equation of the tangent to the circle at the point \( (x_1, y_1) = (1, -1) \) can be given by the formula:
\[ (x - x_1)(x_1 - h) + (y - y_1)(y_1 - k) = 0 \]
Where \( (h, k) \) is the center of the circle \( \left( \frac{4}{3}, \frac{1}{3} \right) \).

Substituting the values:
\[ (x - 1)\left(1 - \frac{4}{3}\right) + (y + 1)\left(-1 - \frac{1}{3}\right) = 0 \]
\[ (x - 1)\left(-\frac{1}{3}\right) + (y + 1)\left(-\frac{4}{3}\right) = 0 \]
Multiplying through by -3 to eliminate the fractions:
\[ 3(x - 1) + 4(y + 1) = 0 \]
\[ 3x - 3 + 4y + 4 = 0 \]
\[ 3x + 4y + 1 = 0 \]

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