Find the lengths of the tangents drawn
to the circle ` 2 x ^(2) + 2y ^(2) = 3 ` from the point (-2, 3)

Answer

To find the lengths of the tangents drawn to the circle \( 2x^2 + 2y^2 = 3 \) from the point \( (-2, 3) \), we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form.
The given equation of the circle is:
\[ 2x^2 + 2y^2 = 3 \]
Dividing the entire equation by 2, we get:
\[ x^2 + y^2 = \frac{3}{2} \]
This represents a circle with center at \( (0, 0) \) and radius \( r = \sqrt{\frac{3}{2}} \).

### Step 2: Identify the coordinates of the external point.
The point from which the tangents are drawn is given as \( P(-2, 3) \).

### Step 3: Use the formula for the length of the tangent from a point to a circle.
The length of the tangent \( PT \) from a point \( P(x_1, y_1) \) to a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ PT = \sqrt{x_1^2 + y_1^2 + 2gx + 2fy + c} \]
For our circle \( x^2 + y^2 - \frac{3}{2} = 0 \), we can identify:
- \( g = 0 \)
- \( f = 0 \)
- \( c = -\frac{3}{2} \)

### Step 4: Substitute the point coordinates into the formula.
Substituting \( P(-2, 3) \) into the formula:
\[ PT = \sqrt{(-2)^2 + (3)^2 + 2(0)(-2) + 2(0)(3) - \frac{3}{2}} \]
Calculating each term:
- \( (-2)^2 = 4 \)
- \( (3)^2 = 9 \)
- \( 2(0)(-2) = 0 \)
- \( 2(0)(3) = 0 \)

So we have:
\[ PT = \sqrt{4 + 9 + 0 + 0 - \frac{3}{2}} = \sqrt{13 - \frac{3}{2}} \]

### Step 5: Simplify the expression under the square root.
To simplify \( 13 - \frac{3}{2} \):
Convert \( 13 \) into a fraction:
\[ 13 = \frac{26}{2} \]
Thus,
\[ PT = \sqrt{\frac{26}{2} - \frac{3}{2}} = \sqrt{\frac{26 - 3}{2}} = \sqrt{\frac{23}{2}} \]

### Final Answer:
The length of the tangents drawn from the point \( (-2, 3) \) to the circle is:
\[ PT = \sqrt{\frac{23}{2}} \]

---