`x^(2)+y^(2)+2lambdax +5=0` and `x^(2)+y^(2)+2lambday+5=0` are the equations of two circles. P is any point on the line `x- y =0` from which the lengths of tangents to the two circles are `t_(1)` and `t_(2)`. If `t_(1) = 3`, then `t_(2)` is

Options

A. `3//2`
B. 3
C. 6
D. none

Answer

To solve the problem, we need to find the lengths of the tangents from a point \( P \) on the line \( x - y = 0 \) to two circles defined by the equations:

1. \( x^2 + y^2 + 2\lambda x + 5 = 0 \) (Circle 1)
2. \( x^2 + y^2 + 2\lambda y + 5 = 0 \) (Circle 2)

Given that the length of the tangent from point \( P \) to Circle 1 is \( t_1 = 3 \), we need to find the length of the tangent to Circle 2, denoted as \( t_2 \).

### Step 1: Identify the point \( P \)
Since \( P \) lies on the line \( x - y = 0 \), we can express \( P \) as \( (a, a) \).

### Step 2: Length of tangent to Circle 1
The length of the tangent from point \( P(a, a) \) to Circle 1 can be calculated using the formula:
\[ t_1 = \sqrt{S_1} \]
where \( S_1 \) is obtained by substituting \( (a, a) \) into the equation of Circle 1.

Substituting \( x = a \) and \( y = a \) into the equation of Circle 1:
\[ S_1 = a^2 + a^2 + 2\lambda a + 5 = 2a^2 + 2\lambda a + 5 \]
Thus, we have:
\[ t_1 = \sqrt{2a^2 + 2\lambda a + 5} \]

Given \( t_1 = 3 \):
\[ 3 = \sqrt{2a^2 + 2\lambda a + 5} \]
Squaring both sides:
\[ 9 = 2a^2 + 2\lambda a + 5 \]
Rearranging gives:
\[ 2a^2 + 2\lambda a - 4 = 0 \]
Dividing the entire equation by 2:
\[ a^2 + \lambda a - 2 = 0 \tag{1} \]

### Step 3: Length of tangent to Circle 2
Now, we calculate the length of the tangent from point \( P(a, a) \) to Circle 2:
\[ t_2 = \sqrt{S_2} \]
where \( S_2 \) is obtained by substituting \( (a, a) \) into the equation of Circle 2.

Substituting \( x = a \) and \( y = a \) into the equation of Circle 2:
\[ S_2 = a^2 + a^2 + 2\lambda a + 5 = 2a^2 + 2\lambda a + 5 \]
Thus, we have:
\[ t_2 = \sqrt{2a^2 + 2\lambda a + 5} \]

### Step 4: Relate \( t_1 \) and \( t_2 \)
From the previous steps, we see that:
\[ t_1 = t_2 \]
This means that the lengths of the tangents from point \( P \) to both circles are equal. Since we know \( t_1 = 3 \), we conclude that:
\[ t_2 = 3 \]

### Final Answer
Thus, the length of the tangent to Circle 2 is:
\[ \boxed{3} \]