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

Options

A. `sqrt(3)`
B. `sqrt((3)/(2))`
C. `2sqrt(3)`
D. none

Answer

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

### Step 1: Rewrite the Circle Equation
First, we simplify the equation of the circle by dividing every term by 2:

\[ x^2 + y^2 + 3x - 3y + \frac{3}{2} = 0 \]

### Step 2: Identify the Center and Radius
Next, we need to rewrite the equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\). We will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + 3x \rightarrow (x + \frac{3}{2})^2 - \frac{9}{4} \]

For \(y\):
\[ y^2 - 3y \rightarrow (y - \frac{3}{2})^2 - \frac{9}{4} \]

Now substituting back into the equation:

\[ (x + \frac{3}{2})^2 - \frac{9}{4} + (y - \frac{3}{2})^2 - \frac{9}{4} + \frac{3}{2} = 0 \]

Combining the constants:
\[ (x + \frac{3}{2})^2 + (y - \frac{3}{2})^2 - \frac{9}{4} - \frac{9}{4} + \frac{6}{4} = 0 \]
\[ (x + \frac{3}{2})^2 + (y - \frac{3}{2})^2 - \frac{12}{4} = 0 \]
\[ (x + \frac{3}{2})^2 + (y - \frac{3}{2})^2 = 3 \]

Thus, the center of the circle is \((- \frac{3}{2}, \frac{3}{2})\) and the radius \(r = \sqrt{3}\).

### Step 3: Use the Length of Tangent Formula
The formula for the length of the tangent \(L\) from a point \((x_1, y_1)\) to a circle centered at \((h, k)\) with radius \(r\) is given by:

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

Substituting the values:
- Point \((x_1, y_1) = (1, 2)\)
- Center \((h, k) = (-\frac{3}{2}, \frac{3}{2})\)
- Radius \(r = \sqrt{3}\)

Calculate \(L\):

\[ L = \sqrt{\left(1 - (-\frac{3}{2})\right)^2 + \left(2 - \frac{3}{2}\right)^2 - (\sqrt{3})^2} \]
\[ = \sqrt{\left(1 + \frac{3}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 3} \]
\[ = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 3} \]
\[ = \sqrt{\frac{25}{4} + \frac{1}{4} - 3} \]
\[ = \sqrt{\frac{26}{4} - \frac{12}{4}} \]
\[ = \sqrt{\frac{14}{4}} = \sqrt{\frac{7}{2}} \]

### Final Answer
The length of the tangent from the point (1, 2) to the circle is:

\[ L = \sqrt{\frac{7}{2}} \]

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