If the length of the tangent from (1,2) to the circle `x^(2)+y^2+x+y-4=0 and 3x^(2)+3y^(2)-x+y+lambda=0` are in the ratio 4:3 then `lambda=`

Options

A. `-55/4`
B. `55/4`
C. `7/4`
D. `-7/4`

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the lengths of the tangents from the point \( (1, 2) \) to the two given circles are in the ratio \( 4:3 \).

### Step 1: Identify the equations of the circles
The equations of the circles are:
1. \( x^2 + y^2 + x + y - 4 = 0 \)
2. \( 3x^2 + 3y^2 - x + y + \lambda = 0 \)

### Step 2: Rewrite the first circle in standard form
The first circle can be rewritten as:
\[ x^2 + y^2 + x + y = 4 \]
Completing the square for \( x \) and \( y \):
\[ (x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 = 4 + \frac{1}{4} + \frac{1}{4} = \frac{17}{4} \]
Thus, the center is \( (-\frac{1}{2}, -\frac{1}{2}) \) and the radius is \( r_1 = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \).

### Step 3: Calculate the length of the tangent to the first circle
Using the formula for the length of the tangent from a point \( (x_1, y_1) \) to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]
For the first circle:
- \( g = \frac{1}{2}, f = \frac{1}{2}, c = -4 \)
- \( (x_1, y_1) = (1, 2) \)

Calculating:
\[ L_1 = \sqrt{1^2 + 2^2 + 2 \cdot \frac{1}{2} \cdot 1 + 2 \cdot \frac{1}{2} \cdot 2 - 4} \]
\[ = \sqrt{1 + 4 + 1 + 2 - 4} = \sqrt{4} = 2 \]

### Step 4: Rewrite the second circle in standard form
The second circle can be rewritten as:
\[ x^2 + y^2 - \frac{1}{3}x + \frac{1}{3}y + \frac{\lambda}{3} = 0 \]
Completing the square:
\[ \left(x - \frac{1}{6}\right)^2 + \left(y + \frac{1}{6}\right)^2 = r_2^2 \]
where \( r_2^2 = \frac{1}{36} + \frac{1}{36} - \frac{\lambda}{3} \).

### Step 5: Calculate the length of the tangent to the second circle
Using the same formula:
\[ L_2 = \sqrt{1^2 + 2^2 - \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 2 + \frac{\lambda}{3}} \]
Calculating:
\[ = \sqrt{1 + 4 - \frac{1}{3} + \frac{2}{3} + \frac{\lambda}{3}} = \sqrt{5 + \frac{\lambda}{3} - \frac{1}{3}} = \sqrt{5 + \frac{\lambda - 1}{3}} \]

### Step 6: Set up the ratio of the lengths
Given that the lengths are in the ratio \( 4:3 \):
\[ \frac{L_1}{L_2} = \frac{4}{3} \]
Substituting the lengths:
\[ \frac{2}{\sqrt{5 + \frac{\lambda - 1}{3}}} = \frac{4}{3} \]

### Step 7: Cross-multiply and solve for \( \lambda \)
Cross-multiplying gives:
\[ 2 \cdot 3 = 4 \cdot \sqrt{5 + \frac{\lambda - 1}{3}} \]
\[ 6 = 4 \sqrt{5 + \frac{\lambda - 1}{3}} \]
Squaring both sides:
\[ 36 = 16 \left(5 + \frac{\lambda - 1}{3}\right) \]
\[ 36 = 80 + \frac{16(\lambda - 1)}{3} \]
Rearranging:
\[ 36 - 80 = \frac{16(\lambda - 1)}{3} \]
\[ -44 = \frac{16(\lambda - 1)}{3} \]
Multiplying both sides by 3:
\[ -132 = 16(\lambda - 1) \]
Dividing by 16:
\[ \lambda - 1 = -\frac{132}{16} = -\frac{33}{4} \]
Thus:
\[ \lambda = 1 - \frac{33}{4} = -\frac{29}{4} \]

### Final Answer
\[ \lambda = -\frac{29}{4} \]