Find the length of the tangents drawn from the point (3,-4) to the circle
`2x^(2)+2y^(2)-7x-9y-13=0`.

Answer

To find the length of the tangents drawn from the point (3, -4) to the circle given by the equation \(2x^2 + 2y^2 - 7x - 9y - 13 = 0\), we can follow these steps:

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

\[ x^2 + y^2 - \frac{7}{2}x - \frac{9}{2}y - \frac{13}{2} = 0 \]

### Step 2: Identify the Coefficients
The standard form of a circle is given by:

\[ x^2 + y^2 + Dx + Ey + F = 0 \]

From our equation, we can identify:
- \(D = -\frac{7}{2}\)
- \(E = -\frac{9}{2}\)
- \(F = -\frac{13}{2}\)

### Step 3: Calculate \(S_1\)
The length of the tangent from an external point \((x_1, y_1)\) to the circle can be calculated using the formula:

\[ L = \sqrt{S_1} \]

where

\[ S_1 = x_1^2 + y_1^2 + Dx_1 + Ey_1 + F \]

Substituting the point \((3, -4)\) into the equation:

\[ S_1 = 3^2 + (-4)^2 + \left(-\frac{7}{2}\right)(3) + \left(-\frac{9}{2}\right)(-4) + \left(-\frac{13}{2}\right) \]

### Step 4: Calculate Each Term
Calculating each term step by step:

1. \(3^2 = 9\)
2. \((-4)^2 = 16\)
3. \(-\frac{7}{2} \times 3 = -\frac{21}{2}\)
4. \(-\frac{9}{2} \times -4 = \frac{36}{2} = 18\)
5. \(-\frac{13}{2} = -\frac{13}{2}\)

### Step 5: Combine the Terms
Now, we combine all these results:

\[ S_1 = 9 + 16 - \frac{21}{2} + 18 - \frac{13}{2} \]

### Step 6: Convert to a Common Denominator
To combine these, we convert them to a common denominator of 2:

\[ S_1 = \frac{18}{2} + \frac{32}{2} - \frac{21}{2} + \frac{36}{2} - \frac{13}{2} \]

### Step 7: Simplify
Now, we can simplify:

\[ S_1 = \frac{18 + 32 - 21 + 36 - 13}{2} = \frac{52}{2} = 26 \]

### Step 8: Calculate the Length of the Tangent
Finally, we find the length of the tangent:

\[ L = \sqrt{S_1} = \sqrt{26} \]

### Final Answer
Thus, the length of the tangents drawn from the point (3, -4) to the circle is:

\[ \sqrt{26} \]

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