The tangent to the circle `C_(1) : x^(2) +y^(2) -2x -1=0` at the point ( 2,1) cuts off a chord of length 4 from a circle `C_(2)` whose centre is`( 3,-2)` . The radius of `C_(2)`is `:`

Options

A. `sqrt(2)`
B. ` sqrt( 6)`
C. 3
D. 2

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Find the equation of the tangent to circle \( C_1 \) at the point \( (2, 1) \)

The equation of the circle \( C_1 \) is given by:
\[ x^2 + y^2 - 2x - 1 = 0 \]
We can rewrite this in standard form:
\[ (x - 1)^2 + (y - 0)^2 = 2 \]
This shows that the center of circle \( C_1 \) is \( (1, 0) \) and the radius is \( \sqrt{2} \).

The formula for the tangent to a circle \( (x_1, y_1) \) is:
\[ (x - x_1)(x_1 - a) + (y - y_1)(y_1 - b) = 0 \]
Substituting \( (x_1, y_1) = (2, 1) \) into the tangent formula gives:
\[ (x - 2)(2 - 1) + (y - 1)(1 - 0) = 0 \]
Simplifying this, we get:
\[ x + y - 3 = 0 \]
Thus, the equation of the tangent line is:
\[ x + y = 3 \]

### Step 2: Determine the distance from the center of circle \( C_2 \) to the tangent line

The center of circle \( C_2 \) is given as \( (3, -2) \). We need to find the distance \( d \) from this point to the line \( x + y = 3 \).

The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( x + y - 3 = 0 \), we have \( A = 1, B = 1, C = -3 \). Substituting \( (x_0, y_0) = (3, -2) \):
\[ d = \frac{|1(3) + 1(-2) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|3 - 2 - 3|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Step 3: Relate the chord length to the radius of circle \( C_2 \)

The length \( l \) of the chord cut off by the tangent line is given as 4. The relationship between the radius \( r \) of circle \( C_2 \), the distance \( d \) from the center to the chord, and the chord length \( l \) is given by:
\[ r^2 = \left(\frac{l}{2}\right)^2 + d^2 \]
Substituting \( l = 4 \) and \( d = \sqrt{2} \):
\[ r^2 = \left(\frac{4}{2}\right)^2 + (\sqrt{2})^2 = 2^2 + 2 = 4 + 2 = 6 \]
Thus, we find:
\[ r = \sqrt{6} \]

### Final Answer
The radius of circle \( C_2 \) is:
\[ \sqrt{6} \]

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