The length of the tangent from the point (1,2) to the circle `2x^(2)+2y^(2)+6x-8y+3=0` is ………. .

Answer

To find the length of the tangent from the point (1,2) to the circle given by the equation \(2x^2 + 2y^2 + 6x - 8y + 3 = 0\), 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 can divide the entire equation by 2 to simplify it:

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

### Step 2: Identify the Center and Radius
The standard form of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. We can find the center and radius using the following formulas:

- Center: \((-g, -f)\) where \(g\) and \(f\) are the coefficients of \(x\) and \(y\) respectively.
- Radius: \(r = \sqrt{g^2 + f^2 - c}\)

From the equation \(x^2 + y^2 + 3x - 4y + \frac{3}{2} = 0\):
- \(g = \frac{3}{2}\)
- \(f = -2\)
- \(c = \frac{3}{2}\)

Calculating the center:
\[ \text{Center} = \left(-\frac{3}{2}, 2\right) \]

Calculating the radius:
\[ r = \sqrt{\left(\frac{3}{2}\right)^2 + (-2)^2 - \frac{3}{2}} = \sqrt{\frac{9}{4} + 4 - \frac{3}{2}} = \sqrt{\frac{9}{4} + \frac{16}{4} - \frac{6}{4}} = \sqrt{\frac{19}{4}} = \frac{\sqrt{19}}{2} \]

### Step 3: Calculate the Length of the Tangent
The length of the tangent from a point \((x_1, y_1)\) to a circle with center \((h, k)\) and radius \(r\) is given by the formula:

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

Substituting the values:
- Point \((x_1, y_1) = (1, 2)\)
- Center \((h, k) = \left(-\frac{3}{2}, 2\right)\)
- Radius \(r = \frac{\sqrt{19}}{2}\)

Calculating \(L\):
\[ L = \sqrt{\left(1 - \left(-\frac{3}{2}\right)\right)^2 + \left(2 - 2\right)^2 - \left(\frac{\sqrt{19}}{2}\right)^2} \]
\[ = \sqrt{\left(1 + \frac{3}{2}\right)^2 + 0 - \frac{19}{4}} = \sqrt{\left(\frac{5}{2}\right)^2 - \frac{19}{4}} = \sqrt{\frac{25}{4} - \frac{19}{4}} = \sqrt{\frac{6}{4}} = \sqrt{\frac{3}{2}} \]

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

\[ \frac{\sqrt{3}}{\sqrt{2}} \text{ or } \frac{\sqrt{6}}{2} \]

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