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 (5, 1) to the given circle defined by the equation \(x^2 + y^2 + 6x - 4y - 3 = 0\), we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \(x^2 + y^2 + 6x - 4y - 3 = 0\), we can identify:
- \(2g = 6 \Rightarrow g = 3\)
- \(2f = -4 \Rightarrow f = -2\)
- \(c = -3\)

### Step 2: Use the formula for the length of the tangent
The formula for the length of the tangent \(L\) from a point \((x_1, y_1)\) to a circle is:
\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]
Here, \((x_1, y_1) = (5, 1)\).

### Step 3: Substitute the values into the formula
Substituting \(x_1 = 5\), \(y_1 = 1\), \(g = 3\), \(f = -2\), and \(c = -3\) into the formula:
\[ L = \sqrt{5^2 + 1^2 + 2(3)(5) + 2(-2)(1) - 3} \]

### Step 4: Calculate each term
Calculating each term:
- \(5^2 = 25\)
- \(1^2 = 1\)
- \(2(3)(5) = 30\)
- \(2(-2)(1) = -4\)

Now, substituting these values back into the equation:
\[ L = \sqrt{25 + 1 + 30 - 4 - 3} \]

### Step 5: Simplify the expression
Now, simplifying the expression inside the square root:
\[ L = \sqrt{25 + 1 + 30 - 4 - 3} = \sqrt{49} \]

### Step 6: Find the final value of \(L\)
Calculating the square root:
\[ L = 7 \]

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

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