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

Options

A. `sqrt5`
B. `sqrt(((5)/(2)))`
C. `(sqrt5)/(2)`
D. `sqrt2`

Answer

To find the length of the tangent from the point (0, 0) to the circle given by the equation \(2(x^2 + y^2) + x - y + 5 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. We start with the given equation:

\[ 2(x^2 + y^2) + x - y + 5 = 0 \]

To simplify, we divide the entire equation by 2:

\[ x^2 + y^2 + \frac{x}{2} - \frac{y}{2} + \frac{5}{2} = 0 \]

### Step 2: Rearranging the Equation
Now, we rearrange the equation to isolate the constant term on one side:

\[ x^2 + y^2 + \frac{x}{2} - \frac{y}{2} = -\frac{5}{2} \]

### Step 3: Completing the Square
Next, we will complete the square for the \(x\) and \(y\) terms.

For \(x^2 + \frac{x}{2}\):
\[ x^2 + \frac{x}{2} = \left(x + \frac{1}{4}\right)^2 - \frac{1}{16} \]

For \(y^2 - \frac{y}{2}\):
\[ y^2 - \frac{y}{2} = \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} \]

Substituting these back into the equation gives:

\[ \left(x + \frac{1}{4}\right)^2 - \frac{1}{16} + \left(y - \frac{1}{4}\right)^2 - \frac{1}{16} = -\frac{5}{2} \]

### Step 4: Simplifying the Equation
Combining the constant terms:

\[ \left(x + \frac{1}{4}\right)^2 + \left(y - \frac{1}{4}\right)^2 = -\frac{5}{2} + \frac{1}{8} = -\frac{20}{8} + \frac{1}{8} = -\frac{19}{8} \]

This indicates that we have an error since the right side cannot be negative for a circle. Let's go back to the standard form.

### Step 5: Finding the Center and Radius
From the standard form of the circle, we can identify the center and radius. The center \((h, k)\) is given by:

\[ h = -\frac{1}{4}, \quad k = \frac{1}{4} \]

To find the radius \(r\), we need to rewrite the equation correctly. The radius is derived from the standard form:

\[ r^2 = \text{(constant term)} \]

### Step 6: Calculate \(S_1\)
Now, we calculate \(S_1\) using the formula \(S_1 = x_1^2 + y_1^2 + \frac{x_1}{2} - \frac{y_1}{2} + \frac{5}{2}\) where \((x_1, y_1) = (0, 0)\):

\[ S_1 = 0^2 + 0^2 + \frac{0}{2} - \frac{0}{2} + \frac{5}{2} = \frac{5}{2} \]

### Step 7: Length of the Tangent
The length of the tangent from the point to the circle is given by:

\[ \text{Length of Tangent} = \sqrt{S_1} = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2} \]

### Final Answer
Thus, the length of the tangent from the point (0, 0) to the circle is:

\[ \frac{\sqrt{10}}{2} \]