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

Options

A. `sqrt(13//2)`
B. `3`
C. 2
D. 1

Answer

To find the length of the tangent from the point (1, 1) to the circle given by the equation \(2x^2 + 2y^2 + 5x + 3y + 1 = 0\), 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:
\[ 2x^2 + 2y^2 + 5x + 3y + 1 = 0 \]
We can divide the entire equation by 2 to simplify it:
\[ x^2 + y^2 + \frac{5}{2}x + \frac{3}{2}y + \frac{1}{2} = 0 \]

### Step 2: Identify Coefficients
Now, we can identify the coefficients \(g\), \(f\), and \(c\) from the standard form of the circle equation:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we have:
- \(2g = \frac{5}{2} \Rightarrow g = \frac{5}{4}\)
- \(2f = \frac{3}{2} \Rightarrow f = \frac{3}{4}\)
- \(c = \frac{1}{2}\)

### Step 3: Find the Center and Radius
The center of the circle \((h, k)\) is given by:
\[ (-g, -f) = \left(-\frac{5}{4}, -\frac{3}{4}\right) \]
The radius \(r\) is calculated using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
Substituting the values:
\[ r = \sqrt{\left(\frac{5}{4}\right)^2 + \left(\frac{3}{4}\right)^2 - \frac{1}{2}} = \sqrt{\frac{25}{16} + \frac{9}{16} - \frac{8}{16}} = \sqrt{\frac{26}{16}} = \frac{\sqrt{26}}{4} \]

### Step 4: Calculate the Distance from the Center to the Point
Next, we need to calculate the distance \(OP\) from the center of the circle to the point \(P(1, 1)\):
\[ OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{\left(1 + \frac{5}{4}\right)^2 + \left(1 + \frac{3}{4}\right)^2} \]
Calculating each term:
\[ = \sqrt{\left(\frac{9}{4}\right)^2 + \left(\frac{7}{4}\right)^2} = \sqrt{\frac{81}{16} + \frac{49}{16}} = \sqrt{\frac{130}{16}} = \frac{\sqrt{130}}{4} \]

### Step 5: Use Pythagorean Theorem to Find the Length of the Tangent
Using the Pythagorean theorem in triangle \(OAP\):
\[ OP^2 = OA^2 + PA^2 \]
Where \(OA\) is the radius and \(PA\) is the length of the tangent we want to find:
\[ \left(\frac{\sqrt{130}}{4}\right)^2 = \left(\frac{\sqrt{26}}{4}\right)^2 + PA^2 \]
This simplifies to:
\[ \frac{130}{16} = \frac{26}{16} + PA^2 \]
Rearranging gives:
\[ PA^2 = \frac{130}{16} - \frac{26}{16} = \frac{104}{16} = \frac{13}{2} \]
Thus, taking the square root:
\[ PA = \sqrt{\frac{13}{2}} = \frac{\sqrt{13}}{\sqrt{2}} = \frac{\sqrt{13}}{2} \]

### Final Answer
The length of the tangent from the point (1, 1) to the circle is:
\[ \frac{\sqrt{13}}{2} \]