`x^2+y^2+2lambdax+5=0` and `x^2+y^2+2lambday+5=0` are two circles. P is a point on the line x-y=0. If PA and PB are the lengths of the tangents from P to the two cricles and PA=3, then PB=

Options

A. 1
B. 3
C. 8
D. 5

Answer

To solve the problem, we need to analyze the two circles given 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)

We also know that point \( P \) lies on the line \( x - y = 0 \), which means \( P \) can be represented as \( (a, a) \) for some value \( a \).

### Step 1: Find the Radical Axis of the Two Circles

The radical axis of two circles is given by the equation \( S_1 - S_2 = 0 \), where \( S_1 \) and \( S_2 \) are the equations of the circles.

Subtracting the equations of the circles:
\[ (x^2 + y^2 + 2\lambda x + 5) - (x^2 + y^2 + 2\lambda y + 5) = 0 \]

This simplifies to:
\[ 2\lambda x - 2\lambda y = 0 \]

Factoring out \( 2\lambda \):
\[ 2\lambda (x - y) = 0 \]

Since \( \lambda \) is a constant and not equal to zero, we have:
\[ x - y = 0 \]

This means the radical axis is the line \( x - y = 0 \).

### Step 2: Understand the Implication of the Radical Axis

The radical axis is the locus of points from which tangents drawn to both circles have equal lengths. Since point \( P \) lies on the radical axis, the lengths of the tangents from \( P \) to both circles will be equal.

### Step 3: Given Information About Tangents

We are given that the length of the tangent from point \( P \) to Circle 1, denoted as \( PA \), is 3:
\[ PA = 3 \]

Since \( P \) lies on the radical axis, we have:
\[ PB = PA \]

Thus, we can conclude:
\[ PB = 3 \]

### Final Answer

Therefore, the length of the tangent from point \( P \) to Circle 2 is:
\[ PB = 3 \]

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