The length of the tangent to the circle `x^(2)+y^(2)-2x-y-7=0` from (-1, -3), is

Options

A. `sqrt(8)`
B. `2sqrt(2)`
C. 4
D. 8

Explanation

To find the length of the tangent from the point (-1, -3) to the circle given by the equation \(x^2 + y^2 - 2x - y - 7 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - y - 7 = 0 \]
We can rearrange this to:
\[ x^2 - 2x + y^2 - y = 7 \]
Next, we complete the square for both \(x\) and \(y\).

### Step 2: Complete the Square
For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
For \(y^2 - y\):
\[ y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4} \]
Now substitute these back into the equation:
\[ (x - 1)^2 - 1 + (y - \frac{1}{2})^2 - \frac{1}{4} = 7 \]
Simplifying this gives:
\[ (x - 1)^2 + (y - \frac{1}{2})^2 = 7 + 1 + \frac{1}{4} = \frac{29}{4} \]
So the center of the circle is \((1, \frac{1}{2})\) and the radius \(r\) is \(\sqrt{\frac{29}{4}} = \frac{\sqrt{29}}{2}\).

### Step 3: Use the Length of Tangent Formula
The length of the tangent \(L\) from a point \((x_1, y_1)\) to a circle with center \((h, k)\) and radius \(r\) is given by:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]
Here, \((x_1, y_1) = (-1, -3)\), \((h, k) = (1, \frac{1}{2})\), and \(r = \frac{\sqrt{29}}{2}\).

### Step 4: Calculate the Length of the Tangent
Now, substituting the values:
\[ L = \sqrt{((-1) - 1)^2 + ((-3) - \frac{1}{2})^2 - \left(\frac{\sqrt{29}}{2}\right)^2} \]
Calculating each term:
\[ (-1 - 1)^2 = (-2)^2 = 4 \]
\[ (-3 - \frac{1}{2})^2 = (-\frac{7}{2})^2 = \frac{49}{4} \]
\[ \left(\frac{\sqrt{29}}{2}\right)^2 = \frac{29}{4} \]
Now substituting these into the length formula:
\[ L = \sqrt{4 + \frac{49}{4} - \frac{29}{4}} = \sqrt{4 + \frac{20}{4}} = \sqrt{4 + 5} = \sqrt{9} = 3 \]

### Final Answer
Thus, the length of the tangent from the point (-1, -3) to the circle is \(3\).