The length of tangent from (5,1) to the circle `x^(2)+y^(2)+6x-4y-3=0` is

Options

A. 81
B. 29
C. 7
D. 21

Answer

To find the length of the tangent from the point \( P(5, 1) \) to the circle defined by the equation \( x^2 + y^2 + 6x - 4y - 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. The given equation is:

\[ x^2 + y^2 + 6x - 4y - 3 = 0 \]

We can group the \( x \) and \( y \) terms:

\[ (x^2 + 6x) + (y^2 - 4y) = 3 \]

### Step 2: Complete the Square
Next, we complete the square for both \( x \) and \( y \).

For \( x^2 + 6x \):
\[ x^2 + 6x = (x + 3)^2 - 9 \]

For \( y^2 - 4y \):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:

\[ ((x + 3)^2 - 9) + ((y - 2)^2 - 4) = 3 \]

Simplifying this, we have:

\[ (x + 3)^2 + (y - 2)^2 - 13 = 3 \]

\[ (x + 3)^2 + (y - 2)^2 = 16 \]

### Step 3: Identify the Center and Radius
From the standard form \( (x + 3)^2 + (y - 2)^2 = 16 \), we can identify the center and radius of the circle:

- Center \( C(-3, 2) \)
- Radius \( r = \sqrt{16} = 4 \)

### Step 4: Use the Length of Tangent Formula
The length of the tangent \( L \) from a point \( P(x_1, y_1) \) to a circle with center \( C(h, k) \) and radius \( r \) is given by:

\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

Substituting \( P(5, 1) \) and \( C(-3, 2) \):

- \( x_1 = 5 \)
- \( y_1 = 1 \)
- \( h = -3 \)
- \( k = 2 \)
- \( r = 4 \)

### Step 5: Calculate the Length of the Tangent
Now we can calculate \( L \):

\[ L = \sqrt{(5 - (-3))^2 + (1 - 2)^2 - 4^2} \]

Calculating each term:

\[ L = \sqrt{(5 + 3)^2 + (1 - 2)^2 - 16} \]
\[ L = \sqrt{(8)^2 + (-1)^2 - 16} \]
\[ L = \sqrt{64 + 1 - 16} \]
\[ L = \sqrt{49} \]
\[ L = 7 \]

### Final Answer
The length of the tangent from the point \( (5, 1) \) to the circle is \( 7 \).

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