Let `2 x^2 + y^2 - 3xy = 0` be the equation of pair of tangents drawn from the origin to a circle of radius 3, with center in the first quadrant. If A is the point of contact. Find OA

Explanation

To solve the problem, we need to find the distance \( OA \), where \( O \) is the origin and \( A \) is the point of contact of the tangents drawn from the origin to the circle. The equation given is \( 2x^2 + y^2 - 3xy = 0 \).

### Step 1: Identify the coefficients
The equation \( 2x^2 + y^2 - 3xy = 0 \) can be compared with the general form of the equation of a pair of straight lines:
\[ ax^2 + by^2 + 2hxy = 0 \]
From this, we identify:
- \( a = 2 \)
- \( b = 1 \)
- \( h = -\frac{3}{2} \)

### Step 2: Calculate \( \tan 2\theta \)
Using the formula for \( \tan 2\theta \):
\[ \tan 2\theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
we first calculate \( h^2 - ab \):
\[ h^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \]
\[ ab = 2 \cdot 1 = 2 \]
Thus,
\[ h^2 - ab = \frac{9}{4} - 2 = \frac{9}{4} - \frac{8}{4} = \frac{1}{4} \]
Now substituting into the formula for \( \tan 2\theta \):
\[ \tan 2\theta = \frac{2\sqrt{\frac{1}{4}}}{2 + 1} = \frac{2 \cdot \frac{1}{2}}{3} = \frac{1}{3} \]

### Step 3: Use the double angle formula for tangent
We know:
\[ \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \]
Setting this equal to \( \frac{1}{3} \):
\[ \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{1}{3} \]
Cross-multiplying gives:
\[ 6\tan \theta = 1 - \tan^2 \theta \]
Rearranging leads to:
\[ \tan^2 \theta + 6\tan \theta - 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 6, c = -1 \):
\[ \tan \theta = \frac{-6 \pm \sqrt{36 + 4}}{2} = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2} = -3 \pm \sqrt{10} \]
Since we are in the first quadrant, we take:
\[ \tan \theta = -3 + \sqrt{10} \]

### Step 5: Find \( OA \)
In triangle \( OAC \), where \( OA \) is the base, \( AC \) is the height (which is the radius of the circle, \( 3 \)), and \( OC \) is the hypotenuse:
\[ \tan \theta = \frac{AC}{OA} = \frac{3}{OA} \]
Thus,
\[ OA = \frac{3}{\tan \theta} = \frac{3}{-3 + \sqrt{10}} \]
To rationalize:
\[ OA = \frac{3(-3 - \sqrt{10})}{(-3 + \sqrt{10})(-3 - \sqrt{10})} = \frac{-9 - 3\sqrt{10}}{9 - 10} = -(-9 - 3\sqrt{10}) = 9 + 3\sqrt{10} \]

### Final Answer
Thus, the distance \( OA \) is:
\[ OA = 9 + 3\sqrt{10} \]